# Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.10 | Set 2

**Evaluate the following limits:**

**Question 16.**

**Solution:**

We have,

=

=

=

Asand, we get,

=

= 2

**Question 17.**

**Solution:**

We have,

=

=

=

Asand, we get,

= log e × 1

= 1

**Question 18.**

**Solution:**

We have,

=

=

=

=

=

Asand, we get,

= 1 −

=

**Question 19.**

**Solution:**

We have,

=

=

=

Let h = x/a − 1. We get,

=

We know,. So, we have,

=

=

**Question 20.**

**Solution:**

We have,

=

=

=

=

=

=

We know,. So, we have,

=

=

**Question 21.**

**Solution:**

We have,

=

=

=

=

We know,. So, we have,

=

**Question 22.**

**Solution:**

We have,

=

=

=

=

We know,. So, we have,

=

**Question 23.**

**Solution:**

We have,

=

=

=

=

=

=

We know,. So, we have,

=

=

**Question 24.**

**Solution:**

We have,

=

=

=

=

As, we get,

= log 8 − log 2

=

= log 4

**Question 25.**

**Solution:**

We have,

=

=

=

Asand, we get,

= (log 2) × 2

= 2 log 2

= log 4

**Question 26.**

**Solution:**

We have,

=

=

=

=

=

=

We know,. So, we have,

=

=

**Question 27.**

**Solution:**

We have,

=

=

=

=

We know,and. So, we get,

= 1

**Question 28. **

**Solution:**

We have,

=

=

We know,. So, we have,

=

= a

^{0}× log a= log a

**Question 29.**

**Solution:**

We have,

=

=

=

=

=

As numerator and denominator are both zero for x = 0, therefore limit cannot exist.

**Question 30. **

**Solution:**

We have,

=

Let x = h + 5. We get,

=

=

=

We know,. So, we have,

= e

^{5}× log e= e

^{5}

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