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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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