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

  • Last Updated : 30 Apr, 2021

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

Solution:

We have, Limx→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, Limx→0(x/|x|) does not exist.

Question 2. Find k so that Limx→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 Limx→0(1/x) does not exist.

Solution:

We have to show that Limx→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, Limx→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 limx→0 f(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 limx→0 f(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, limx→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 limx→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 limx→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, limx→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 limx→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 limx→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, limx→0f(x) does not exist.

Question 7. Find limx→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 limx→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, limx→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 limx→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 limx→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, limx→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 limx→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 limx→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 limx→1f(x) does not exist.

Question 9. Find limx→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 limx→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, limx→1f(x) does not exist.

Question 10. Evaluate limx→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 limx→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, limx→0f(x) does not exist.

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).

Solution:

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

\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 – an)

= 0

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

Now, put x = a

= (a – a1)(a – a2)……..(a – an)

Hence, limx→af(x) = (a – a1)(a – a2)……..(a – an)

Question 12. Find limx→1+[1/(x – 1)].

Solution:

We have to find limx→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, limx→1+[1/(x – 1)] = ∞

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

Solution:

We have, limx→0+[1/3x]

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

= Limh→0+[1/3(0+h)]

= Limh→0+[1/(3h)]

= ∞

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

Solution:

We have, limx→-8+[2x/(x + 8)]

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

= limx→0+[2(-8 + h)/(-8 + h + 8)]

= Limh→0+[(2h – 16)/(h)]

= -∞

Question 13(v). Evaluate the following one-sided limits: limx→0+[2/x1/5]

Solution:

We have, limx→0+[2/x1/5]

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

= Limh→0+[2/(0 + h)1/5]

= ∞

Question 13(vi). Evaluate the following one-sided limits: limx→(π/2)[tanx]

Solution:

We have, limx→(π/2)[tanx]

Let x = 0 – h, where h = 0.

= limh→0[tan(π/2 – h)]

= limx→0[cot h]

= ∞

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

Solution:

We have, limx→(-π/2)+[secx]

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

= limh→0+[secx(-π/2 + h)]

= limh→0+[cosec h]

= ∞

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

Solution:

We have, limx→0-[x2 – 3x + 2/x3 – 2x2]

= Limx→0-[(x – 1)(x – 2)/x2(x – 2)]

= Limx→0-[(x – 1)/x2]

Let x = 0 – h, where h = 0.

= Limh→0-[(0 – h – 1)/(0 – h)2]

= -∞

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

Solution:

We have, limx→-2+[(x2 – 1)/(2x + 4)]

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

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

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

= (4 – 1)/0

= ∞

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

Solution:

We have, limx→0-[2 – cotx]

Let x = 0 – h, where h = 0.

= Limh→0-[2 – cot(0 – h)]

= Limh→0-[2 + cot(h)]

= 2 + ∞

= ∞

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

Solution:



We have, limx→0-[1 + cosecx]

Let x = 0 – h, where h = 0.

= Limh→0-[1 + cosec(0 – h)]

= Limh→0-[1 – cosec(h)]

= 1 – ∞

= -∞

Question 14. Show that Limx→0e-1/x does not exist.

Solution:

Let, f(x) = Limx→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, Limx→0e-1/x does not exist.

Question 15(i). Find Limx→2[x]  

Solution:

We have, Limx→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, Limx→2[x] does not exist.

Question 15(ii). Find Limx→5/2[x]  

Solution:

We have, Limx→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, Limx→5/2[x] = 2

Question 15(iii). Find Limx→1[x]  

Solution:

We have, Limx→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, Limx→1[x] does not exist.

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

Solution:

We have, \lim_{x\to a^+}[x]

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

= Limh→0-[(a + h)]

= a

Also, \lim_{x\to1^-}[x]

Let x = 1 – h, where h = 0.

= Limh→0[(1 – h)]

= 0

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

Solution:

We have to show Limx→2+(x/[x]) ≠ Limx→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.

= Limh→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.

= Limh→0+[(2 + h)/|[2 + h]]

= 2/2

= 1

Hence, Left-hand limit≠Right-hand limit

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

Solution:

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

Let x = 3 – h, where h = 0.

= Limh→0-[(3 – h)/|[3 – h]]

= 3/2

Also, \lim_{x\to3^+}\frac{x}{[x]}

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

= Limh→0+[(3 + h)/|[3 + h]]

= 3/3

= 1

Hence, Left-hand limit≠Right-hand limit

Question 19. Find Limx→5/2[x]  

Solution:

We have to find Limx→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.

= Limh→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.

= Limh→0+[(5/2+h)]

= 2

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

Question 20. Evaluate Limx→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 Limx→2f(x)

So for that 

First we find left-hand limit:



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

Let x = 2 – h, where h = 0.

= Limh→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.

= Limh→0-[3(2 + h) – 5]

= 6 – 5

= 1

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

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

Solution:

Let, f(x) = Limx→0sin(1/x)

First we find left-hand limit:

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

Let x = 0 – h, where h = 0.

= Limh→0sin[1/(0 – h)]

= -Limh→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, Limx→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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