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

### 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:

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, Limx→0(x/|x|) does not exist.

### Question 2. Find k so that Limx→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 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:

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, 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.

Solution:

We have,

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:

=

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, limx→0 f(x) does not exist.

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

Solution:

We have,

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

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

Solution:

We have,

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

### Question 7. Find limx→3f(x), where

Solution:

We have,

And we have to find limx→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, limx→3f(x) = 4

### Question 8(i). If , Find limx→0f(x).

Solution:

We have,

And we have to find limx→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, limx→0f(x) = 3

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

Solution:

We have,

And we have to find limx→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 limx→1f(x) does not exist.

### Question 9. Find limx→1f(x) Where

Solution:

We have,

And we have to find limx→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, limx→1f(x) does not exist.

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

Solution:

We have,

And we have to find limx→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, 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)

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)]

=

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

=

=

= ∞

Hence, limx→1+[1/(x – 1)] = ∞

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

Solution:

We have,

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

=

= -∞

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

Solution:

We have,

Let x = 2 – h, where h = 0.

= ∞

### 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:

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, 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:

=

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, 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:

=

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, 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:

=

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, Limx→1[x] does not exist.

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

Solution:

We have,

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

= Limh→0-[(a + h)]

= a

Also,

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

Let x = 3 – h, where h = 0.

= Limh→0-[(3 – h)/|[3 – h]]

= 3/2

Also,

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:

Let x = 5/2 – h, where h = 0.

= Limh→0-[(5/2 – h)]

= 2

Now we find right-hand limit:

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

Solution:

We have,

We have to find Limx→2f(x)

So for that

First we find left-hand limit:

Let x = 2 – h, where h = 0.

= Limh→0-{(2 – h) – [2 – h]}

= 2 – 1

= 1

Now we find right-hand limit:

=

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:

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 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

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