# NCERT Solutions class-11 Maths Exercise 13.1

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Exercise 13.1

Evaluate the following limits in Exercises 1 to 22.

1.

Ans. 3 + 3 = 6

2.

Ans.

3.

Ans.

4.

Ans.

5.

Ans.

6.

Ans.

Putting as

7.

Ans.

=

=

8.

Ans.

=

=

=

9.

Ans.

10.

Ans.

=

=

=

= = 1 + 1 = 2

11.

Ans.

=

= = 1

12.

Ans. =

=

= =

13.

Ans.

=

= =

14.

Ans.

=

=

=

=

15.

Ans.

Putting as

=

= =

= =

16.

Ans. =

17.

Ans. =

= =

=

= =

18.

Ans.

=

=

=

=

19.

Ans. =

= = = 0

20.

Ans.

Dividing numerator and denominator by

=

= = = 1

21.

Ans. Given:

= =

= =

= = 0

22.

Ans. Given:

Putting as

= =

= =

=

23. Find and where

Ans. Given:

Now =

And = 3 (1 + 1) =

24. Find where

Ans. Given:

L.H.L.

Putting as

=

= = 0

R.H.L.

Putting as

=

= =

25. Evaluate where

Ans. Given:

L.H.L.

Putting as

= = =

R.H.L.

Putting as

= = =

Here, L.H.L. R.H.L.

Therefore, this limit does not exist at

26. Find where

Ans. Given:

L.H.L.

Putting as

= = =

R.H.L.

Putting as

= = =

Here, L.H.L. R.H.L.

Therefore, this limit does not exist at

27. Find where

Ans. Given:

L.H.L.

Putting as

=

= = 0

R.H.L.

Putting as

=

= = 0

Here, L.H.L. = R.H.L.

Therefore, this limit exists at and

28. Suppose and if what are possible values of and ?

Ans. Given: and

and……… (i)

Now

Putting as

= ……….(ii)

Again

Putting as

= ……….(ii)

Putting values from eq. (ii) and (iii) in eq. (i), we get

and

On solving these equation, we get and

29. Let be fixed real numbers and define a function What is ? For some compute

Ans. Given:

Now

=

= = 0

Also

=

30. If for what values of does exists?

Ans. Given:

exists for all

Now L.H.L.

Putting as

=

= = 0 + 1 = 1

Also R.H.L.

Putting as

=

= =

Here, L.H.L. R.H.L.

Therefore, this limit does not exist.

31. If the function satisfies valuate

Ans. Since

32. If for what integer and does both and exist?

Ans. Both and exist.

and

Now

= =

And

= =

……….(i)

Therefore, for exists we need

Again

= =

And

=

=

Therefore, exists for any integral value of and

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