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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,