You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

## 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.

\(\lim _{x \rightarrow 1}\)(x^{2} + 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 x_{0} for which \(\lim _{x \rightarrow x_{0}}\) f(x) exists.

Question 16.

Solution:

Question 17.

Solution:

Limit exists except at x_{0} = 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 a_{1}, a_{2} …………… a_{n} be fixed real numbers such that f(x) = (x – a_{1}) , (x – a_{2}), ………. (x – a_{n}) what \(\lim _{x \rightarrow a}\) f(x) For a ≠ a_{1}, a_{2}, ………… a_{n} compute \(\lim _{x \rightarrow a}\) f(x)

Solution:

We have,