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.    Question 2.    Question 3.    Question 4.    Question 5.    Question 6.    Question 7.  From the graph the value of the function at x = 3 is y = f(3) = 1 Question 8.   From the graph the value of the function at x = 1 is y = f(1) = 3   Question 9.   Question 10.    From the graph the value of the function is y = f(1) = 3   Question 11.   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.  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 $$\lim _{x \rightarrow 1}$$ 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 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.  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 x0 for which $$\lim _{x \rightarrow x_{0}}$$ f (x) exists.

Question 16.  At x = 4 , the curve does not exist. Hence, except at x0 = 4 , the limit of f(x) exists. Question 17.  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 x0 = 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 (ii) f(-2) = 0
f(2) = 0
$$\lim _{x \rightarrow-2}$$ f(x) = 0
$$\lim _{x \rightarrow 2}$$ f(x) does not exist.  Question 19.
Write a brief description of the meaning of the notation $$\lim _{x \rightarrow 8}$$ f(x) = 25
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?
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.
$$\lim _{x \rightarrow 2}$$ 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.   Verify the existence of $$\lim _{x \rightarrow 0}$$ f(x), where  