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

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:

= 

Let x = 0 – h, where h = 0

= 

= 

= -1

Now we find right-hand limit:

= 

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

= 

= 

= 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 

Solution:

We have, 

Now first we find left-hand limit:

= 

Let x = 2 – h, where h= 0.

=

= [2(2 – 0) + 3]

= 7

Now we find right-hand limit:

= 

Let x = 2 + h, where h = 0

= 

= (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:

= 

Let x = 0 – h, where h = 0.

= 

= 

= -∞

Now we find right-hand limit:

=

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

= 

= 

= ∞

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 . Show that lim_x→0 f(x) does not exist.

Solution:

We have, 

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

So for that 

First we find left-hand limit:

=

Let x = 0 – h, where h = 0

=

=

=

= 3

Now we find right-hand limit:

=

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

=

=

=

= 1

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

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

Solution:

We have, 

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

So for that 

First we find left-hand limit:

=

Let x = 0 – h, where h = 0.

=

=

= -1

Now we find right-hand limit:

=

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

=

=

= 1

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

Question 6. Let , Prove that lim_x→0f(x) does not exist.

Solution:

We have,

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

So for that 

First we find left-hand limit:

=

Let x = 0 – h, where h = 0.

=

=

= -4

Now we find right-hand limit:

=

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

=

=

= 5

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

Question 7. Find lim_x→3f(x), where 

Solution:

We have, 

And we have to find lim_x→3f(x)

So for that 

First we find left-hand limit:

=

Let x = 3 – h, where h = 0.

= 

= 

= 4

Now we find right-hand limit:

=

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

=

= 4

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

Hence, lim_x→3f(x) = 4

Question 8(i). If , Find lim_x→0f(x).

Solution:

We have, 

And we have to find lim_x→0f(x)

So for that 

First we find left-hand limit:

=

Let x = 0 – h, where h = 0.

= 

= 

= 3

Now we find right-hand limit:

=

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

=

=

= 3

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

Hence, lim_x→0f(x) = 3

Question 8(ii). If , Find lim_x→1f(x).

Solution:

We have, 

And we have to find lim_x→1f(x)

So for that 

First we find left-hand limit:

=

Let x = 1 – h, where h = 0.

= 

= 

= 5

Now we find right-hand limit:

=

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

=

= 

= 6

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

Question 9. Find lim_x→1f(x) Where 

Solution:

We have, 

And we have to find lim_x→1f(x)

So for that 

First we find left-hand limit:

=

Let x = 1 – h, where h = 0.

= 

=

= 0

Now we find right-hand limit:

=

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

= 

= 

= -2

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

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

Solution:

We have, 

And we have to find lim_x→0f(x)

So for that 

First we find left-hand limit:

=

Let x = 0 – h, where h = 0.

= 

= 

= -1

Now we find right-hand limit:

=

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

= 

= 

= 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)

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)]

=

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

=

=

= ∞

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, 

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

=

= 

= 

= -∞

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

Solution:

We have,

Let x = 2 – h, where h = 0.

= 

= 

= 

= ∞

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/x does not exist.

Solution:

Let, f(x) = Lim_x→0e^-1/x

So for that 

First we find left-hand limit:

= 

Let x = 0 – h, where h = 0.

= 

=

= e^∞

= ∞

Now we find right-hand limit:

=

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

= 

=

= 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:

=

Let x = 2 – h, where h = 0.

= 

= 1

Now we find right-hand limit:

= 

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

= 

= 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:

=

Let x = 5/2 – h, where h = 0.

= 

= 2

Now we find right-hand limit:

=

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

= 

= 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:

=

Let x = 1 – h, where h = 0.

= 

= 0

Now we find right-hand limit:

= 

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

= 

= 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, 

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

= Lim_h→0-[(a + h)]

= a

Also, 

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, , 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, , 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,  Where [] is Greatest Integer Function

Let x = 3 – h, where h = 0.

= Lim_h→0-[(3 – h)/|[3 – h]]

= 3/2

Also, 

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:

= 

Let x = 5/2 – h, where h = 0.

= Lim_h→0-[(5/2 – h)]

= 2

Now we find right-hand limit:

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 

Solution:

We have, 

We have to find Lim_x→2f(x)

So for that 

First we find left-hand limit:

= 

Let x = 2 – h, where h = 0.

= Lim_h→0-{(2 – h) – [2 – h]}

= 2 – 1

= 1

Now we find right-hand limit:

=

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:

= 

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 and if lim _x→π​/2 f(x) = f(π/2), find the value of k.

Solution:

We have 

First we find left-hand limit:

= 

Let x = π/2 – h, where h = 0.

=

= k cos(π/2 – π/2)/π

= k/π

Now we find right-hand limit:

Let x = π/2 + h, where h = 0.

=

= 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