Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.1

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Question 1. Show that Lim_x→0(x/|x|) does not exist.

Solution:

We have, Lim_x→0(x/|x|)

Now first we find left-hand limit:

= $\lim_{x\to0^-}\frac{x}{|x|}$

Let x = 0 – h, where h = 0

= $\lim_{h\to0^-}(\frac{(0 - h)}{|0 - h|})$

= $\lim_{h\to0^-}(\frac{(- h)}{h})$

= -1

Now we find right-hand limit:

= $\lim_{x\to0^+}\frac{x}{|x|}$

So, let x = 0 + h, where h = 0

= $\lim_{h\to0^+}(\frac{(0 + h)}{|0 + h|})$

= $\lim_{h\to0^+}(\frac{h}{h})$

= 1

Left-hand limit ≠ Right-hand limit

So, Lim_x→0(x/|x|) does not exist.

Question 2. Find k so that Lim_x→0f(x), where $f(x)= \begin{cases} 2x+3, \hspace{0.2cm}x\le2\\ x+k,\hspace{0.2cm}x>2 \end{cases}$

Solution:

We have, $f(x)= \begin{cases} 2x+3, \hspace{0.2cm}x\le2\\ x+k,\hspace{0.2cm}x>2 \end{cases}$

Now first we find left-hand limit:

= $\lim_{x\to0^-}(2x+3)$

Let x = 2 – h, where h= 0.

= $\lim_{h\to0^-}(2(2-h)+3)$

= [2(2 – 0) + 3]

= 7

Now we find right-hand limit:

$=\lim_{x\to0^+}f(x)$

= $\lim_{x\to0^+}(x+k)$

Let x = 2 + h, where h = 0

= $\lim_{h\to0^+}((2+h)+k)$

= (2 + 0) + k

= (2 + k)

Here, Left-hand limit = Right-hand limit, so limit exists

So, (2 + k) = 7

k = 5

Question 3. Show that Lim_x→0(1/x) does not exist.

Solution:

We have to show that Lim_x→0(1/x) does not exists

So for that

First we find left-hand limit:

= $\lim_{x\to0^-}\frac{1}{x}$

Let x = 0 – h, where h = 0.

= $\lim_{h\to0^-}(\frac{1}{(0-h)})$

= $-\lim_{h\to0^-}(\frac{1}{h})$

= -∞

Now we find right-hand limit:

= $\lim_{x\to0^+}\frac{1}{x}$

Let x = 0 + h, where h = 0.

= $\lim_{h\to0^+}(\frac{1}{|(0+h)|})$

= $\lim_{h\to0^+}(\frac{1}{h})$

= ∞

Here, Left-hand limit ≠ Right-hand limit, so, Lim_x→0(1/x) does not exist.

Question 4. Let f(x) be a function defined by $f(x)= \begin{cases} \frac{3x}{(|x|+2x)}, \hspace{0.2cm}x ≠ 0\\ 0,\hspace{0.2cm}x=0 \end{cases}$ . Show that lim_x→0f(x) does not exist.

Solution:

We have, $f(x)= \begin{cases} \frac{3x}{(|x|+2x)}, \hspace{0.2cm}x ≠ 0\\ 0,\hspace{0.2cm}x=0 \end{cases}$

According to the question we have to show that lim_x→0f(x) does not exist.

So for that

First we find left-hand limit:

= $\lim_{x\to0^-}\frac{3x}{|x|+2x}$

Let x = 0 – h, where h = 0

= $\lim_{h\to0^-}\frac{3(0-h)}{|0-h|+2(0-h)}$

= $\lim_{h\to0^-}(\frac{-3h}{h - 2h})$

= $\lim_{h\to0^-}(\frac{-3}{-1})$

= 3

Now we find right-hand limit:

= $\lim_{x\to0^+}\frac{3x}{|x|+2x}$

Let x = 0 + h, where h = 0.

= $\lim_{h\to0^+}\frac{3(0+h)}{|0+h|+2(0+h)}$

= $\lim_{h\to0^+}(\frac{3h}{3h})$

= $\lim_{h\to0^+}(\frac{3}{3})$

= 1

Here, Left-hand limit ≠ Right-hand limit, so, lim_x→0 f(x) does not exist.

Question 5. Let $f(x)= \begin{cases} x+1, \hspace{0.2cm}if \ x >0\\ x-1,\hspace{0.2cm}if \ x<0 \end{cases}$ , Prove that lim_x→0f(x) does not exist.

Solution:

We have, $f(x)= \begin{cases} x+1, \hspace{0.2cm}if \ x >0\\ x-1,\hspace{0.2cm}if \ x<0 \end{cases}$

And we have to prove that lim_x→0f(x) does not exist.

So for that

First we find left-hand limit:

= $\lim_{x\to0^-}(x-1)$

Let x = 0 – h, where h = 0.

= $\lim_{h\to0^-}((0-h)-1)$

= $\lim_{h\to0^-}(0-h-1)$

= -1

Now we find right-hand limit:

= $\lim_{x\to0^+}(x+1)$

Let x = 0 + h, where h = 0.

= $\lim_{h\to0^+}((0+h)+1)$

= $\lim_{h\to0^+}(0+h+1)$

= 1

Here, Left-hand limit ≠ Right-hand limit, so, lim_x→0f(x) does not exist.

Question 6. Let $f(x)= \begin{cases} x+5, \hspace{0.2cm}if \ x >0\\ x-4,\hspace{0.2cm}if \ x<0 \end{cases}$ , Prove that lim_x→0f(x) does not exist.

Solution:

We have, $f(x)= \begin{cases} x+5, \hspace{0.2cm}if \ x >0\\ x-4,\hspace{0.2cm}if \ x<0 \end{cases}$

And we have to prove that lim_x→0f(x) does not exist.

So for that

First we find left-hand limit:

= $\lim_{x\to0^-}(x-4)$

Let x = 0 – h, where h = 0.

= $\lim_{h\to0^-}((0-h)-4)$

= $\lim_{h\to0^-}(0-h-4)$

= -4

Now we find right-hand limit:

= $\lim_{x\to0^+}(x+5)$

Let x = 0 + h, where h = 0.

= $\lim_{h\to0^+}((0+h)+5)$

= $\lim_{h\to0^+}(0+h+5)$

= 5

Here, Left-hand limit ≠ Right-hand limit, so, lim_x→0f(x) does not exist.

Question 7. Find lim_x→3f(x), where $f(x)= \begin{cases} 4, \hspace{0.2cm}if \ x >3\\ x+1,\hspace{0.2cm}if \ x<3 \end{cases}$

Solution:

We have, $f(x)= \begin{cases} 4, \hspace{0.2cm}if \ x >3\\ x+1,\hspace{0.2cm}if \ x<3 \end{cases}$

And we have to find lim_x→3f(x)

So for that

First we find left-hand limit:

= $\lim_{x\to3^-}(x+1)$

Let x = 3 – h, where h = 0.

= $\lim_{h\to0^-}((3-h)+1)$

= $\lim_{h\to0^-}(4-h)$

= 4

Now we find right-hand limit:

= $\lim_{x\to3^+}4$

Let x = 3 + h, where h = 0.

= $\lim_{h\to0^+}4$

= 4

Here, Left-hand limit = Right-hand limit,

Hence, lim_x→3f(x) = 4

Question 8(i). If $f(x)= \begin{cases} 2x+3, \hspace{0.2cm}if \ x\le0\\ 3(x+1),\hspace{0.2cm}if \ x>0 \end{cases}$ , Find lim_x→0f(x).

Solution:

We have, $f(x)= \begin{cases} 2x+3, \hspace{0.2cm}if \ x\le0\\ 3(x+1),\hspace{0.2cm}if \ x>0 \end{cases}$

And we have to find lim_x→0f(x)

So for that

First we find left-hand limit:

= $\lim_{x\to0^-}(2x+3)$

Let x = 0 – h, where h = 0.

= $\lim_{h\to0^-}(2(0-h)+3)$

= $\lim_{h\to0^-}(3-2h)$

= 3

Now we find right-hand limit:

= $\lim_{x\to0^+}3(x+1)$

Let x = 0 + h, where h = 0.

= $\lim_{h\to0^+}3((0+h)+1)$

= $\lim_{h\to0^+}(3+3h)$

= 3

Here, Left-hand limit = Right-hand limit,

Hence, lim_x→0f(x) = 3

Question 8(ii). If $f(x)= \begin{cases} 2x+3, \hspace{0.2cm}if \ x\le0\\ 3(x+1),\hspace{0.2cm}if \ x>0 \end{cases}$ , Find lim_x→1f(x).

Solution:

We have, $f(x)= \begin{cases} 2x+3, \hspace{0.2cm}if \ x\le0\\ 3(x+1),\hspace{0.2cm}if \ x>0 \end{cases}$

And we have to find lim_x→1f(x)

So for that

First we find left-hand limit:

= $\lim_{x\to1^-}(2x+3)$

Let x = 1 – h, where h = 0.

= $\lim_{h\to0^-}(2(1-h)+3)$

= $\lim_{h\to0^-}(5-2h)$

= 5

Now we find right-hand limit:

= $\lim_{x\to1^+}3(x+1)$

Let x = 1 + h, where h = 0.

= $\lim_{h\to0^+}(3((1+h)+1))$

= $\lim_{h\to0^+}(6+3h)$

= 6

Here, Left-hand limit ≠ Right-hand limit, so lim_x→1f(x) does not exist.

Question 9. Find lim_x→1f(x) Where $f(x)= \begin{cases} x^2-1, \hspace{0.2cm}if \ x\le1\\ -x^2-1,\hspace{0.2cm}if \ x>1 \end{cases}$

Solution:

We have, $f(x)= \begin{cases} x^2-1, \hspace{0.2cm}if \ x\le1\\ -x^2-1,\hspace{0.2cm}if \ x>1 \end{cases}$

And we have to find lim_x→1f(x)

So for that

First we find left-hand limit:

= $\lim_{x\to1^-}(x^2-1)$

Let x = 1 – h, where h = 0.

= $\lim_{h\to0^-}((1-h)^2-1)$

= $\lim_{h\to0^-}(1-2h-h^2-1)$

= 0

Now we find right-hand limit:

= $\lim_{x\to1^+}(-x^2-1)$

Let x = 1 + h, where h = 0.

= $\lim_{h\to0^+}(-(1+h)^2-1)$

= $\lim_{h\to0^+}(-1-2h-h^2-1)$

= -2

Here, Left-hand limit ≠ Right-hand limit, so, lim_x→1f(x) does not exist.

Question 10. Evaluate lim_x→0f(x), where $f(x)= \begin{cases} \frac{|x|}{x}, \hspace{0.2cm}if \ x≠1\\ 0,\hspace{0.2cm}if \ x=1 \end{cases}$

Solution:

We have, $f(x)= \begin{cases} \frac{|x|}{x}, \hspace{0.2cm}if \ x≠1\\ 0,\hspace{0.2cm}if \ x=1 \end{cases}$

And we have to find lim_x→0f(x)

So for that

First we find left-hand limit:

= $\lim_{x\to0^-}\frac{|x|}{x}$

Let x = 0 – h, where h = 0.

= $\lim_{h\to0^-}\frac{|0-h|}{0-h}$

= $\lim_{h\to0^-}(\frac{h}{-h})$

= -1

Now we find right-hand limit:

= $\lim_{x\to0^+}\frac{|x|}{x}$

Let x = 0 + h, where h = 0.

= $\lim_{h\to0^+}\frac{|0+h|}{0+h}$

= $\lim_{h\to0^+}(\frac{h}{h})$

= 1

Here, Left-hand limit ≠ Right-hand limit, so, lim_x→0f(x) does not exist.

Question 11. Let a₁, a₂,……….a_n be fixed real number such that f(x) = (x – a₁)(x – a₂)……..(x-a_n). What is lim_x→a1f(x)? Compute lim_x→af(x).

Solution:

We have, f(x) = (x – a₁)(x – a₂)……..(x – a_n)

$\lim_{x→a_1}f(x)=\lim_{x→a_1}[(x-a_1)(x-a_2)........(x-a_n)]$

Now, put x = a₁

= (a₁– a₁)(a₁– a₂)……..(a₁– a_n)

= 0

Now, lim_x→af(x) = lim_x→a[(x – a₁)(x – a₂)……..(x – a_n)]

Now, put x = a

= (a – a₁)(a – a₂)……..(a – a_n)

Hence, lim_x→af(x) = (a – a₁)(a – a₂)……..(a – a_n)

Question 12. Find lim_x→1⁺[1/(x – 1)].

Solution:

We have to find lim_x→1⁺[1/(x – 1)]

= $\lim_{x\to1^+}\frac{1}{x-1}$

Let x = 1 + h, where h = 0.

= $\lim_{h\to0^+}\frac{1}{(1+h)-1}$

= $\lim_{h\to0^+}\frac{1}{(h)}$

= ∞

Hence, lim_x→1⁺[1/(x – 1)] = ∞

Question 13(i). Evaluate the following one-sided limits: lim_x→2⁺[(x – 3)/(x²– 4)]

Solution:

We have, $\lim_{x\to2^+}\frac{x-3}{x^2-4}$

Let x = 2 + h, where h = 0.

= $\lim_{h\to0^+}(\frac{(2+h)-3}{(2+h)^2-4})$

= $\lim_{h\to0^+}(\frac{h-1}{4+4h+h^2-4})$

= $\lim_{h\to0^+}(\frac{h-1}{4h+h^2})$

= -∞

Question 13(ii). Evaluate the following one-sided limits: lim_x→2^–[(x – 3)/(x²– 4)]

Solution:

We have, $\lim_{x\to2^-}\frac{x-3}{x^2-4}$

Let x = 2 – h, where h = 0.

= $\lim_{h\to0^-}\frac{(2-h)-3}{(2-h)^2-4}$

= $\lim_{h\to0^-}\frac{(-h-1)}{4-4h+h^2-4}$

= $\lim_{h\to0^-}\frac{(-h-1)}{-4h+h^2}$

= ∞

Question 13(iii). Evaluate the following one-sided limits: lim_x→0⁺[1/3x]

Solution:

We have, lim_x→0⁺[1/3x]

Let x = 0 + h, where h = 0.

= Lim_h→0⁺[1/3(0+h)]

= Lim_h→0⁺[1/(3h)]

= ∞

Question 13(iv). Evaluate the following one-sided limits: lim_x→-8⁺[2x/(x + 8)]

Solution:

We have, lim_x→-8⁺[2x/(x + 8)]

Let x = -8 + h, where h = 0.

= lim_x→0⁺[2(-8 + h)/(-8 + h + 8)]

= Lim_h→0⁺[(2h – 16)/(h)]

= -∞

Question 13(v). Evaluate the following one-sided limits: lim_x→0⁺[2/x^1/5]

Solution:

We have, lim_x→0⁺[2/x^1/5]

Let x = 0 + h, where h = 0.

= Lim_h→0⁺[2/(0 + h)^1/5]

= ∞

Question 13(vi). Evaluate the following one-sided limits: lim_{x→(π/2)^–}[tanx]

Solution:

We have, lim_{x→(π/2)^–}[tanx]

Let x = 0 – h, where h = 0.

= lim_h→0^–[tan(π/2 – h)]

= lim_x→0^–[cot h]

= ∞

Question 13(vii). Evaluate the following one-sided limits: lim_{x→(-π/2)⁺}[secx]

Solution:

We have, lim_{x→(-π/2)⁺}[secx]

Let x = 0 + h, where h = 0.

= lim_h→0⁺[secx(-π/2 + h)]

= lim_h→0⁺[cosec h]

= ∞

Question 13(viii). Evaluate the following one-sided limits: lim_x→0^–[(x²– 3x + 2)/x³– 2x²]

Solution:

We have, lim_x→0-[x²– 3x + 2/x³– 2x²]

= Lim_x→0-[(x – 1)(x – 2)/x²(x – 2)]

= Lim_x→0-[(x – 1)/x²]

Let x = 0 – h, where h = 0.

= Lim_h→0-[(0 – h – 1)/(0 – h)²]

= -∞

Question 13(ix). Evaluate the following one-sided limits: lim_x→-2⁺[(x²– 1)/(2x + 4)]

Solution:

We have, lim_x→-2⁺[(x²– 1)/(2x + 4)]

Let x = -2 + h, where h = 0.

= Lim_h→-0⁺[(-2 + h)²– 1)/2(-2 + h) + 4]

= Lim_h→-0⁺[(-2 + h)²– 1)/(-4 + 4 + h)]

= (4 – 1)/0

= ∞

Question 13(x). Evaluate the following one-sided limits: lim_x→0-[2 – cotx]

Solution:

We have, lim_x→0-[2 – cotx]

Let x = 0 – h, where h = 0.

= Lim_h→0-[2 – cot(0 – h)]

= Lim_h→0-[2 + cot(h)]

= 2 + ∞

= ∞

Question 13(xi). Evaluate the following one-sided limits. lim_x→0-[1 + cosecx]

Solution:

We have, lim_x→0-[1 + cosecx]

Let x = 0 – h, where h = 0.

= Lim_h→0-[1 + cosec(0 – h)]

= Lim_h→0-[1 – cosec(h)]

= 1 – ∞

= -∞

Question 14. Show that Lim_x→0e^-1/xdoes not exist.

Solution:

Let, f(x) = Lim_x→0e^-1/x

So for that

First we find left-hand limit:

= $\lim_{x\to0^-}e^{-\frac{1}{x}}$

Let x = 0 – h, where h = 0.

= $\lim_{h\to0^-}e^{-\frac{1}{0-h}}$

= $\lim_{h\to0^-}e^{\frac{1}{h}}$

= e^∞

= ∞

Now we find right-hand limit:

= $\lim_{x\to0^+}e^{-\frac{1}{x}}$

Let x = 0 + h, where h = 0.

= $\lim_{h\to0^+}e^{-\frac{1}{0 + h}}$

= $\lim_{h\to0^+}e^{-\frac{1}{h}}$

= e^-∞

= 0

Here, Left-hand limit ≠ Right-hand limit, so, Lim_x→0e^-1/x does not exist.

Question 15(i). Find Lim_x→2[x]

Solution:

We have, Lim_x→2[x], where [] is Greatest Integer Function

So for that

First we find left-hand limit:

= $\lim_{x\to0^-}[x]$

Let x = 2 – h, where h = 0.

= $\lim_{h\to0^-}[2 - h]$

= 1

Now we find right-hand limit:

= $\lim_{x\to0^+}[x]$

Let x = 2 + h, where h = 0.

= $\lim_{h\to0^+}[2+h]$

= 2

Here, Left-hand limit ≠ Right-hand limit, so, Lim_x→2[x] does not exist.

Question 15(ii). Find Lim_x→5/2[x]

Solution:

We have, Lim_x→2[x], where [] is Greatest Integer Function

So for that

First we find left-hand limit:

= $\lim_{x\to\frac{5}{2}^-}[x]$

Let x = 5/2 – h, where h = 0.

= $\lim_{h\to0^-}[\frac{5}{2} - h]$

= 2

Now we find right-hand limit:

= $\lim_{x\to\frac{5}{2}^+}[x]$

Let x = 5/2 + h, where h = 0.

= $\lim_{h\to0^+}[\frac{5}{2}+h]$

= 2

Here, Left-hand limit = Right-hand limit, so, Lim_x→5/2[x] = 2

Question 15(iii). Find Lim_x→1[x]

Solution:

We have, Lim_x→1[x], where [] is Greatest Integer Function

So for that

First we find left-hand limit:

= $\lim_{x\to1^-}[x]$

Let x = 1 – h, where h = 0.

= $\lim_{h\to0^-}[1-h]$

= 0

Now we find right-hand limit:

= $\lim_{x\to1^+}[x]$

Let x = 1 + h, where h = 0.

= $\lim_{h\to0^+}[1+h]$

= 1

Here, Left-hand limit = Right-hand limit, so, Lim_x→1[x] does not exist.

Question 16. Prove that Lim_x→a+[x] = [a]. Also prove that Lim_x→1-[x] = 0.

Solution:

We have, $\lim_{x\to a^+}[x]$

Let x = a + h, where h = 0.

= Lim_h→0-[(a + h)]

= a

Also, $\lim_{x\to1^-}[x]$

Let x = 1 – h, where h = 0.

= Lim_h→0[(1 – h)]

= 0

Question 17. Show that Lim_x→2+(x/[x]) ≠ Lim_x→2-(x/[x]).

Solution:

We have to show Lim_x→2+(x/[x]) ≠ Lim_x→2-(x/[x])

So, R.H.L

We have, $\lim_{x\to2^-}\frac{x}{[x]}$ , where [] is greatest Integer Function

Let x = 2 – h, where h = 0.

= Lim_h→0-[(2 – h)/|[2 – h]]

= 2/1

= 2

Now, L.H.L

We have, $\lim_{x\to2^+}\frac{x}{[x]}$ , where [] is greatest Integer Function

Let x = 2 + h, where h = 0.

= Lim_h→0+[(2 + h)/|[2 + h]]

= 2/2

= 1

Hence, Left-hand limit≠Right-hand limit

Question 18. Find Lim_x→3+(x/[x]). Is it equal to Lim_x→3-(x/[x])

Solution:

We have, $\lim_{x\to3^-}\frac{x}{[x]}$ Where [] is Greatest Integer Function

Let x = 3 – h, where h = 0.

= Lim_h→0-[(3 – h)/|[3 – h]]

= 3/2

Also, $\lim_{x\to3^+}\frac{x}{[x]}$

Let x = 3 + h, where h = 0.

= Lim_h→0+[(3 + h)/|[3 + h]]

= 3/3

= 1

Hence, Left-hand limit≠Right-hand limit

Question 19. Find Lim_x→5/2[x]

Solution:

We have to find Lim_x→5/2[x], where [] is Greatest Integer Function

So for that

First we find left-hand limit:

= $\lim_{x\to\frac{5}{2}^-}[x]$

Let x = 5/2 – h, where h = 0.

= Lim_h→0-[(5/2 – h)]

= 2

Now we find right-hand limit:

$=\lim_{x\to\frac{5}{2}^+}[x]$

Let x = 5/2 + h, where h = 0.

= Lim_h→0+[(5/2+h)]

= 2

Hence, Left-hand limit = Right-hand limit, so Lim_x→5/2[x] = 2

Question 20. Evaluate Lim_x→2f(x), where $f(x)= \begin{cases} x-[x], \hspace{0.2cm}x<2\\ 4 , \hspace{0.2cm}x=2\\ x+k,\hspace{0.2cm}x>2 \end{cases}$

Solution:

We have, $f(x)= \begin{cases} x-[x], \hspace{0.2cm}x<2\\ 4 , \hspace{0.2cm}x=2\\ x+k,\hspace{0.2cm}x>2 \end{cases}$

We have to find Lim_x→2f(x)

So for that

First we find left-hand limit:

= $\lim_{x\to0^-}(x-[x])$

Let x = 2 – h, where h = 0.

= Lim_h→0-{(2 – h) – [2 – h]}

= 2 – 1

= 1

Now we find right-hand limit:

= $\lim_{x\to2^+}(3x-5)$

Let x = 2 + h, where h = 0.

= Lim_h→0-[3(2 + h) – 5]

= 6 – 5

= 1

Hence, Left-hand limit = Right-hand limit, so, Lim_x→2f(x) = 1

Question 21. Show that Lim_x→0sin(1/x) does not exist.

Solution:

Let, f(x) = Lim_x→0sin(1/x)

First we find left-hand limit:

= $\lim_{x\to0^-}sin\frac{1}{x}$

Let x = 0 – h, where h = 0.

= Lim_h→0sin[1/(0 – h)]

= -Lim_h→0sin[1/(h)]

An oscillating number lies between -1 to +1.

So left hand limit does not exists.

Similarly, right-hand limit is also oscillating.

So, Lim_x→0sin(1/x) does not exist.

Question 22. Let $f(x)= \begin{cases} \frac{kcosx}{\pi-2x}, \hspace{0.2cm}where \ x \neq \frac{\pi}{2}\\ 3,\hspace{0.2cm}where \ x =\frac{\pi}{2} \end{cases}$ and if lim_x→_π/2 f(x) = f(π/2), find the value of k.

Solution:

We have $f(x)= \begin{cases} \frac{kcosx}{\pi-2x}, \hspace{0.2cm}where \ x \neq \frac{\pi}{2}\\ 3,\hspace{0.2cm}where \ x =\frac{\pi}{2} \end{cases}$

First we find left-hand limit:

= $\lim_{x\to\frac{\pi}{2}^-}\frac{kcosx}{\pi-2x}$

Let x = π/2 – h, where h = 0.

= $\lim_{h\to0^-}\frac{kcos(\frac{\pi}{2}-h)}{\pi-2(\frac{\pi}{2}-h)}$

= k cos(π/2 – π/2)/π

= k/π

Now we find right-hand limit:

$\lim_{x\to\frac{\pi}{2}^+}\frac{kcosx}{\pi-2x}$

Let x = π/2 + h, where h = 0.

= $\lim_{h\to0^+}\frac{kcos(\frac{\pi}{2}+h)}{\pi-2(\frac{\pi}{2}+h)}$

= k cos(π/2 + π/2)/-π

= k/π

Hence, Left-hand limit = Right-hand limit, so

lim_x→_π/2 f(x) = f(π/2)

k/π = 3

k = 3π

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Last Updated : 30 Apr, 2021

Class 11 RD Sharma Solutions - Chapter 28 Introduction to 3D Coordinate Geometry - Exercise 28.3

Class 11 RD Sharma Solutions- Chapter 29 Limits - Exercise 29.2

Chapter 1: Sets

Chapter 2: Relations

Chapter 3: Functions

Chapter 4: Measurement of Angles

Chapter 5: Trigonometric Functions

Chapter 6: Graphs of Trigonometric Functions

Chapter 7: Trigonometric Ratios of Compound Angles

Chapter 8: Transformation Formulae

Chapter 9: Trigonometric Ratios of Multiple and Sub Multiple Angles

Chapter 10: Sine and Cosine Formulae and their Applications

Chapter 11: Trigonometric Equations

Chapter 12: Mathematical Induction

Chapter 13: Complex Numbers

Chapter 15: Linear Inequations

Chapter 16: Permutations

Chapter 17: Combinations

Chapter 18: Binomial Theorem

Chapter 19: Arithmetic Progressions

Chapter 20: Geometric Progressions

Chapter 21: Some Special Series

Chapter 22: Brief Review of Cartesian System of Rectangular Coordinates

Chapter 23: The Straight Lines

Chapter 24: The Circle

Chapter 25: Parabola

Chapter 26: Ellipse

Chapter 27: Hyperbola

Chapter 28: Introduction to 3D Coordinate Geometry

Chapter 29: Limits

Chapter 30: Derivatives

Chapter 31: Mathematical Reasoning

Chapter 32: Statistics

Chapter 33: Probability

Chapter 1: Sets

Chapter 2: Relations

Chapter 3: Functions

Chapter 4: Measurement of Angles

Chapter 5: Trigonometric Functions

Chapter 6: Graphs of Trigonometric Functions

Chapter 7: Trigonometric Ratios of Compound Angles

Chapter 8: Transformation Formulae

Chapter 9: Trigonometric Ratios of Multiple and Sub Multiple Angles

Chapter 10: Sine and Cosine Formulae and their Applications

Chapter 11: Trigonometric Equations

Chapter 12: Mathematical Induction

Chapter 13: Complex Numbers

Chapter 15: Linear Inequations

Chapter 16: Permutations

Chapter 17: Combinations

Chapter 18: Binomial Theorem

Chapter 19: Arithmetic Progressions

Chapter 20: Geometric Progressions

Chapter 21: Some Special Series

Chapter 22: Brief Review of Cartesian System of Rectangular Coordinates

Chapter 23: The Straight Lines

Chapter 24: The Circle

Chapter 25: Parabola

Chapter 26: Ellipse

Chapter 27: Hyperbola

Chapter 28: Introduction to 3D Coordinate Geometry

Chapter 29: Limits

Chapter 30: Derivatives

Chapter 31: Mathematical Reasoning

Chapter 32: Statistics

Chapter 33: Probability

Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.1

Question 1. Show that Limx→0(x/|x|) does not exist.

Question 2. Find k so that Limx→0f(x), where

Question 3. Show that Limx→0(1/x) does not exist.

Question 4. Let f(x) be a function defined by . Show that limx→0 f(x) does not exist.

Question 5. Let , Prove that limx→0f(x) does not exist.

Question 7. Find limx→3f(x), where

Question 8(i). If , Find limx→0f(x).

Question 8(ii). If , Find limx→1f(x).

Question 9. Find limx→1f(x) Where

Question 10. Evaluate limx→0f(x), where

Question 11. Let a1, a2,……….an be fixed real number such that f(x) = (x – a1)(x – a2)……..(x-an). What is limx→a1f(x)? Compute limx→af(x).

Question 12. Find limx→1+[1/(x – 1)].

Question 13(i). Evaluate the following one-sided limits: limx→2+[(x – 3)/(x2 – 4)]

Question 13(ii). Evaluate the following one-sided limits: limx→2–[(x – 3)/(x2 – 4)]

Question 13(iii). Evaluate the following one-sided limits: limx→0+[1/3x]

Question 1. Show that Lim_x→0(x/|x|) does not exist.

Question 2. Find k so that Lim_x→0f(x), where $f(x)= \begin{cases} 2x+3, \hspace{0.2cm}x\le2\\ x+k,\hspace{0.2cm}x>2 \end{cases}$

Question 3. Show that Lim_x→0(1/x) does not exist.

Question 4. Let f(x) be a function defined by $f(x)= \begin{cases} \frac{3x}{(|x|+2x)}, \hspace{0.2cm}x ≠ 0\\ 0,\hspace{0.2cm}x=0 \end{cases}$ . Show that lim_x→0f(x) does not exist.

Question 5. Let $f(x)= \begin{cases} x+1, \hspace{0.2cm}if \ x >0\\ x-1,\hspace{0.2cm}if \ x<0 \end{cases}$ , Prove that lim_x→0f(x) does not exist.

Question 7. Find lim_x→3f(x), where $f(x)= \begin{cases} 4, \hspace{0.2cm}if \ x >3\\ x+1,\hspace{0.2cm}if \ x<3 \end{cases}$

Question 8(i). If $f(x)= \begin{cases} 2x+3, \hspace{0.2cm}if \ x\le0\\ 3(x+1),\hspace{0.2cm}if \ x>0 \end{cases}$ , Find lim_x→0f(x).

Question 8(ii). If $f(x)= \begin{cases} 2x+3, \hspace{0.2cm}if \ x\le0\\ 3(x+1),\hspace{0.2cm}if \ x>0 \end{cases}$ , Find lim_x→1f(x).

Question 9. Find lim_x→1f(x) Where $f(x)= \begin{cases} x^2-1, \hspace{0.2cm}if \ x\le1\\ -x^2-1,\hspace{0.2cm}if \ x>1 \end{cases}$

Question 10. Evaluate lim_x→0f(x), where $f(x)= \begin{cases} \frac{|x|}{x}, \hspace{0.2cm}if \ x≠1\\ 0,\hspace{0.2cm}if \ x=1 \end{cases}$

Question 11. Let a₁, a₂,……….a_n be fixed real number such that f(x) = (x – a₁)(x – a₂)……..(x-a_n). What is lim_x→a1f(x)? Compute lim_x→af(x).

Question 12. Find lim_x→1⁺[1/(x – 1)].

Question 13(i). Evaluate the following one-sided limits: lim_x→2⁺[(x – 3)/(x²– 4)]

Question 13(ii). Evaluate the following one-sided limits: lim_x→2^–[(x – 3)/(x²– 4)]

Question 13(iii). Evaluate the following one-sided limits: lim_x→0⁺[1/3x]

Question 13(iv). Evaluate the following one-sided limits: lim_x→-8⁺[2x/(x + 8)]

Question 13(v). Evaluate the following one-sided limits: lim_x→0⁺[2/x^1/5]

Question 13(vi). Evaluate the following one-sided limits: lim_{x→(π/2)^–}[tanx]

Question 13(vii). Evaluate the following one-sided limits: lim_{x→(-π/2)⁺}[secx]

Question 13(viii). Evaluate the following one-sided limits: lim_x→0^–[(x²– 3x + 2)/x³– 2x²]

Question 13(ix). Evaluate the following one-sided limits: lim_x→-2⁺[(x²– 1)/(2x + 4)]

Question 13(x). Evaluate the following one-sided limits: lim_x→0-[2 – cotx]

Question 13(xi). Evaluate the following one-sided limits. lim_x→0-[1 + cosecx]

Question 14. Show that Lim_x→0e^-1/xdoes not exist.

Question 15(i). Find Lim_x→2[x]

Question 15(ii). Find Lim_x→5/2[x]

Question 15(iii). Find Lim_x→1[x]

Question 16. Prove that Lim_x→a+[x] = [a]. Also prove that Lim_x→1-[x] = 0.

Question 17. Show that Lim_x→2+(x/[x]) ≠ Lim_x→2-(x/[x]).

Question 18. Find Lim_x→3+(x/[x]). Is it equal to Lim_x→3-(x/[x])

Question 19. Find Lim_x→5/2[x]

Question 20. Evaluate Lim_x→2f(x), where $f(x)= \begin{cases} x-[x], \hspace{0.2cm}x<2\\ 4 , \hspace{0.2cm}x=2\\ x+k,\hspace{0.2cm}x>2 \end{cases}$

Question 21. Show that Lim_x→0sin(1/x) does not exist.

Question 22. Let $f(x)= \begin{cases} \frac{kcosx}{\pi-2x}, \hspace{0.2cm}where \ x \neq \frac{\pi}{2}\\ 3,\hspace{0.2cm}where \ x =\frac{\pi}{2} \end{cases}$ and if lim_x→_π/2 f(x) = f(π/2), find the value of k.