Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 9 Limits and Continuity Ex 9.1 Text Book Back Questions and Answers, Notes.

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

In Problems 1 – 6, complete the table using calculator and use the result to estimate the limit.

Question 1.

Answer:

Question 2.

Answer:

Question 3.

Answer:

Question 4.

Answer:

Question 5.

Answer:

Question 6.

Answer:

Question 7.

Answer:

From the graph the value of the function at x = 3 is y = f(3) = 1

Question 8.

Answer:

From the graph the value of the function at x = 1 is y = f(1) = 3

Question 9.

Answer:

Question 10.

Answer:

From the graph the value of the function is y = f(1) = 3

Question 11.

Answer:

From the graph the value of the function at x = 3 the curve does not meet the line x = 3

∴ The value of the function is not defined at the point x = 3.

Hence \(\lim _{x \rightarrow 3} \frac{1}{x-3}\) does not exist at x = 3.

Question 12.

Answer:

From the graph x = 5 curve does not intersect the line x = 5

∴ The value of the function y = f(x) does not exist at x = 5.

Question 13.

\(\lim _{x \rightarrow 1}\) sin πx

Answer:

\(\lim _{x \rightarrow 1}\) sin πx

From the graph x = 1, the curve y = f(x) intersects the line x = 1 at x – axis.

∴ y = f(1) = 0

Hence \(\lim _{x \rightarrow 1}\) sin πx = 0

Question 14.

\(\lim _{x \rightarrow 0}\) sec x

Answer:

To find \(\lim _{x \rightarrow 0}\) sec x

Let y = f(x) = sec x

From the graph at x = 0 the curve intersect the y – axis.

At x = 0 we have y = 1

∴ \(\lim _{x \rightarrow 0}\) sec x = 1

Question 15.

Answer:

y = f(x) = sec x

From the graph at x = \(\frac{\pi}{2}\), the curve does not intersect the line x = \(\frac{\pi}{2}\)

At x = \(\frac{\pi}{2}\), the value of the function y = f(x) does not exist.

Hence 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.

Answer:

At x = 4 , the curve does not exist. Hence, except at x_{0} = 4 , the limit of f(x) exists.

Question 17.

Answer:

From the figure when x = π, y = f(π) = 2. The function is not defined at x = π since sin x lies in the interval [ – 1, 1]

∴ The given function has limits at all points except at x = π

(π, 2) point is not possible since the range of the curve is [- 1 , 1] . Except x_{0} = π, the curve has limits.

Question 18.

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

(i) f(0) is defined

\(\lim _{x \rightarrow 0}\) f(x) = 4

f(2) = 6

\(\lim _{x \rightarrow 2}\) f(x) = 3

Answer:

(ii) f(-2) = 0

f(2) = 0

\(\lim _{x \rightarrow-2}\) f(x) = 0

\(\lim _{x \rightarrow 2}\) f(x) does not exist.

Answer:

Question 19.

Write a brief description of the meaning of the notation \(\lim _{x \rightarrow 8}\) f(x) = 25

Answer:

Given \(\lim _{x \rightarrow 8}\) f(x) = 25

By the definition of limit

∴ f(8^{–}) = f(8^{+}) = 25

Question 20.

If f(2) = 4, can you conclude anything about the limit of f (x) as x approaches 2?

Answer:

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 x approaches 2 is 4, can you conclude anything about f (2)? Explain reasoning.

Answer:

\(\lim _{x \rightarrow 2}\) 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.

Answer:

Question 23.

Verify the existence of \(\lim _{x \rightarrow 0}\) f(x), where

Answer:

From equations (1) and (2) we get

f(1^{–}) ≠ f(1^{+})

∴ The limit of f(x) does not exist.