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## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 9 Limits and Continuity Ex 9.1

In problems 1-6, complete the table using calculate and use the result to estimate the limit.
Question 1.

Solution:

Question 2.

Solution:

∴ Limit is 0.25

Question 3.

Solution:

∴ Limit is 0.288

Question 4.

Solution:

∴ Limit is -0.25

Question 5.

Solution:

∴ Limit is 1

Question 6.

Solution:

∴ Limit is 0

In exercise problems 7-15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
Question 7.
$$\lim _{x \rightarrow 3}$$(4 – x)
Solution:

Limit exist and is equal to 1

Question 8.
$$\lim _{x \rightarrow 1}$$(x2 + 2)
Solution:

Limit exist and is equal to = 3

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Limit does not exist

Question 12.

Solution:
When x → 5, (x – 5) = -(x – 5)
∴ $$\lim _{x \rightarrow 5^{-}} \frac{-(x-5)}{(x-5)}$$ = -1
When x → 5+, (x – 5) = (x – 5)

Question 13.
$$\lim _{x \rightarrow 1}$$ sin(πx)
Solution:

Question 14.
$$\lim _{x \rightarrow 0}$$ (sec x)
Solution:

Question 15.
$$\lim _{x \rightarrow \frac{\pi}{2}}$$ tan x
Solution:

Limit does not exist

Sketch the graph of f, then identify the values of x0 for which $$\lim _{x \rightarrow x_{0}}$$ f(x) exists.
Question 16.

Solution:

Question 17.

Solution:

Limit exists except at x0 = π

Question 18.
Sketch the graph of a function f that satisfies the given values:

Solution:

Question 19.
Write a brief description of the meaning of the notation $$\lim _{x \rightarrow 8}$$ f(x) = 25
Solution:

Question 20.
If f(2) = 4, can you conclude anything about the limit of f(x) as x approaches 2?
Solution:
No, f(x) = 4, It is the value of the function at x = 2
This limit doesn’t exists at x = 2
Since f(2) = 4
It need not imply that $$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)$$
∴ we cannot conclude at x = 2

Question 21.
If the limit of f(x) as z approaches 2 is 4, can you conclude anything about f(2)?
Explain reasoning.
Solution:
$$\lim _{x \rightarrow 2}$$ f(x) 4 , $$\lim _{x \rightarrow 2^{-}}$$ f(x) = $$\lim _{x \rightarrow 2^{+}}$$ f(x) = 4
When x approaches 2 from the left or from the right f(x) approaches 4.
Given that $$\lim _{x \rightarrow 2^{-}}$$ f(x) = $$\lim _{x \rightarrow 2^{+}}$$ f(x) = 4
The existence or non existence at x =2 has no leaving on the existence of the limit of f(x) as x approaches to 2.
∴ We cannot conclude the value of f(2)

Question 22.
Evaluate: $$\lim _{x \rightarrow 3} \frac{x^{2}-9}{x-3}$$ if it exists by finding f(3) and f(3+).
Solution:

Question 23.
Verify the existence of
Solution:

Limit does not exist

### Samacheer Kalvi 11th Maths Solutions Chapter 9 Limits and Continuity Ex 9.1 Additional Questions

Question 1.
Suppose . What are possible values of a and b?
Solution:

Question 2.

Solution:
We have,

Question 3.

Solution:
We have,

Question 4.

Solution:

Question 5.
Let a1, a2 …………… an be fixed real numbers such that f(x) = (x – a1) , (x – a2), ………. (x – an) what $$\lim _{x \rightarrow a}$$ f(x) For a ≠ a1, a2, ………… an compute $$\lim _{x \rightarrow a}$$ f(x)
Solution:
We have,