Exercise 9.1-Additional Questions - Chapter 9 - Limits and Continuity - 11th Maths Guide Samacheer Kalvi Solutions

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Additional Questions
Question 1.
$
f(x)=\left\{\begin{array}{l}
a+b x, x<1 \\
4, x=1 \\
b-a x, x>1
\end{array} \text { and, if } \lim _{x \rightarrow 1} f(x)=f(1) .\right.
$
What are possible values of $a$ and $b$ ?

Solution:
We have, $\quad \lim _{x \rightarrow 1} f(x)=f(1)$
$
\begin{aligned}
& \Leftrightarrow \quad \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1) \\
& \Leftrightarrow \quad \lim _{x \rightarrow 1^{-}} f(x)=f(1) \text { and, } \lim _{x \rightarrow 1^{+}} f(x)=f(1) \\
& \text { ¿. } \quad \lim _{x \rightarrow 0} f(1-h)=4 \text { and, } \lim _{x \rightarrow 0} f(1+h)=4 \\
& \Leftrightarrow \quad \lim _{x \rightarrow 0}\{a+b(1-h)\}=4 \text { and, } \lim _{x \rightarrow 0}\{b-a(1+h)\}=4 \\
& \Leftrightarrow \quad a+b=4 \text { and, } b-a=4 \\
& a=0, b=4 \\
&
\end{aligned}
$
Question 2.
If $f(x)=\left\{\begin{array}{l}2 x+3, x \leq 0 \\ 3(x+1), x>0\end{array}\right.$. Find $\lim _{x \rightarrow 0} f(x)$ and $\lim _{x \rightarrow 1} f(x)$.
Solution:
We have,
$
\begin{aligned}
& f(x)= \begin{cases}2 x+3, & x \leq 0 \\
3(x+1) & x \leq 0\end{cases} \\
& \therefore \quad \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(2 x+3)=2 \times 0+3=3 \\
&
\end{aligned}
$
So, $\lim _{x \rightarrow 0} f(x)$ exists and is equal to 3
$
\begin{aligned}
& \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} 2 x+3=2 \times 1+3=5 \\
& \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}} 3(x+1)=3(1+1)=6 \\
& \lim _{x \rightarrow 1^{-}} f(x) \neq \lim _{x \rightarrow 1^{+}} f(x)
\end{aligned}
$
Hence, $\lim _{x \rightarrow 1} f(x)$ does not exist.

Question 3.
Find $\lim _{x \rightarrow 1} f(x)$, if $f(x)=\left\{\begin{array}{c}x^2-1, x \leq 1 \\ -x^2-1, x>1\end{array}\right.$.
Solution:
We have,
$
\begin{aligned}
& f(x)=\left\{\begin{array}{c}
x^2-1, x \leq 1 \\
-x^2-1, x>1
\end{array}\right. \\
& \therefore \quad \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} x^2-1=1^2-1=0 \\
& \text { and } \\
& \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}-x^2-1=-1-1=-2 \\
& \therefore \quad \lim _{x \rightarrow 1^{-}} f(x) \neq \lim _{x \rightarrow 1^{+}} f(x) \\
&
\end{aligned}
$
Hence, $\lim _{x \rightarrow 1} f(x)$ does not exist.

Question 4.
Evaluate $\lim _{x \rightarrow 0} f(x)$, where $f(x)=\left\{\begin{array}{c}\frac{|x|}{x}, x \neq 0 \\ 0, x=0\end{array}\right.$
Solution:

We have,
$
f(x)=\left\{\begin{array}{c}
\frac{|x|}{x}, x \neq 0 \\
0, x=0
\end{array}\right.
$
$
\begin{array}{ll}
\therefore & \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{|x|}{x}=\lim _{x \rightarrow 0^{-}} \frac{-x}{x}=-1 \\
\text { and, } & \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{|x|}{x}=\lim _{x \rightarrow 0} \frac{x}{x}=1 \\
\therefore & \lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x)
\end{array}
$
Hence, $\lim _{x \rightarrow 1} f(x)$ does not exist.
Question 5.
Let $a_1, a_2 \ldots \ldots \ldots \ldots \ldots a_n$ be fixed real numbers such that $f(x)=\left(x-a_1\right),\left(x-a_2\right), \ldots \ldots \ldots\left(x-a_n\right)$ what $\lim _{x \rightarrow a} f(x)$ For $a \neq a_1, a_2, \ldots \ldots \ldots \ldots a_n$ compute $\lim _{x \rightarrow a} f(x)$
Solution:

We have, we have,
$
\begin{aligned}
\therefore \quad \lim _{x \rightarrow a} f(x) & =\lim _{x \rightarrow 0} f\left(a_1-h\right)=\lim _{x \rightarrow 0}-h\left(a_1-h-a_2\right)\left(a_1-h-a_3\right) \ldots\left(a_1-h-a_n\right) \\
& =0\left(a_1-a_2\right)\left(a_1-a_3\right) \ldots\left(a_1-a_n\right)=0 \\
\lim _{x \rightarrow a_1^{+}} f(x) & =\lim _{x \rightarrow 0}\left(a_1+h-a_1\right)\left(a_1+h-a_2\right)\left(a_1+h-a_3\right) \ldots\left(a_1+h-a_n\right) \\
& =\lim _{x \rightarrow 0} h\left(a_1+h-a_2\right)\left(a_1+h-a_3\right) \ldots\left(a_1+h-a_n\right) \\
& =0\left(a_1-a_2\right)\left(a_1-a_3\right) \ldots\left(a_1-a_n\right)=0 \\
\therefore \quad \lim _{x \rightarrow a} f(x) & =\lim _{x \rightarrow a_1^{+}} f(x)=0
\end{aligned}
$
Hence, $\lim _{x \rightarrow a_1} f(x)=0$.
For any $a \neq a_1, a_2, \ldots a_n$.
$
\begin{aligned}
\lim _{x \rightarrow a} f(x) & =\lim _{x \rightarrow a}\left(x-a_1\right)\left(x-a_2\right)\left(x-a_3\right) \ldots\left(x-a_n\right) \\
& =\left(a-a_1\right)\left(a-a_2\right)\left(a-a_3\right) \ldots\left(a-a_n\right)
\end{aligned}
$