# 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→0}f(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 existsSo 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→0}f(x) does not exist.

**Solution:**

We have,

And we have to prove 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.

=

=

= -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→0}f(x) does not exist.

**Question 6. **Let , Prove that lim_{x→0}f(x) does not exist.

**Solution:**

We have,

And we have to prove 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.

=

=

= -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→0}f(x) does not exist.

### Question 7. Find lim_{x→3}f(x), where

**Solution:**

We have,

And we have to find lim

_{x→3}f(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→3}f(x) = 4

### Question 8(i). If , Find lim_{x→0}f(x).

**Solution:**

We have,

And we have to find lim

_{x→0}f(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→0}f(x) = 3

### Question 8(ii). If , Find lim_{x→1}f(x).

**Solution:**

We have,

And we have to find lim

_{x→1}f(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→1}f(x) does not exist.

### Question 9. Find lim_{x→1}f(x) Where

**Solution:**

We have,

And we have to find lim

_{x→1}f(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→1}f(x) does not exist.

### Question 10. Evaluate lim_{x→0}f(x), where

**Solution:**

We have,

And we have to find lim

_{x→0}f(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→0}f(x) does not exist.

### Question 11. Let a_{1}, a_{2},……….a_{n} be fixed real number such that f(x) = (x – a_{1})(x – a_{2})……..(x-a_{n}). What is lim_{x→a1}f(x)? Compute lim_{x→a}f(x).

**Solution:**

We have, f(x) = (x – a

_{1})(x – a_{2})……..(x – a_{n})Now, put x = a

_{1}= (a

_{1 }– a_{1})(a_{1 }– a_{2})……..(a_{1 }– a_{n})= 0

Now, lim

_{x→a}f(x) = lim_{x→a}[(x – a_{1})(x – a_{2})……..(x – a_{n})]Now, put x = a

= (a – a

_{1})(a – a_{2})……..(a – a_{n})Hence, lim

_{x→a}f(x) = (a – a_{1})(a – a_{2})……..(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)/(x^{2 }– 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)/(x^{2 }– 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–}[(x^{2 }– 3x + 2)/x^{3 }– 2x^{2}]

**Solution:**

We have, lim

_{x→0}-[x^{2 }– 3x + 2/x^{3 }– 2x^{2}]= Lim

_{x→0}-[(x – 1)(x – 2)/x^{2}(x – 2)]= Lim

_{x→0}-[(x – 1)/x^{2}]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+}[(x^{2 }– 1)/(2x + 4)]

**Solution:**

We have, lim

_{x→-2+}[(x^{2 }– 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→0}e^{-1/x }does not exist.

**Solution:**

Let, f(x) = Lim

_{x→0}e^{-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→0}e^{-1/x}does not exist.

### Question 15(i). Find Lim_{x→2}[x]

**Solution:**

We have, Lim

_{x→2}[x], where [] is Greatest Integer FunctionSo 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 FunctionSo 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 FunctionSo 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 FunctionSo 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→2}f(x), where

**Solution:**

We have,

We have to find Lim

_{x→2}f(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→2}f(x) = 1

### Question 21. Show that Lim_{x→0}sin(1/x) does not exist.

**Solution:**

Let, f(x) = Lim

_{x→0}sin(1/x)First we find left-hand limit:

=

Let x = 0 – h, where h = 0.

= Lim

_{h→0}sin[1/(0 – h)]= -Lim

_{h→0}sin[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→0}sin(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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