Exercise 9.3-Additional Questions - Chapter 9 - Limits and Continuity - 11th Maths Guide Samacheer Kalvi Solutions

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Additional Problems
Question 1.
Evaluate $\lim _{x \rightarrow \infty} \sqrt{x}(\sqrt{x+c}-\sqrt{x})$
Solution:
The given expression is of the form $\infty-\infty$. So we first write it in the rational form $\frac{f(x)}{g(x)}$. So that it reduces to either $\frac{0}{0}$ form or $\frac{\infty}{\infty}$ form.

$
\begin{aligned}
\therefore \quad \lim _{x \rightarrow \infty} \sqrt{x}\{\sqrt{x+c}-\sqrt{x}\} & =\lim _{x \rightarrow \infty} \frac{\sqrt{x}\{\sqrt{x+c}-\sqrt{x}\}\{\sqrt{x+c}+\sqrt{x}\}}{\{\sqrt{x+c}+\sqrt{x}\}} \\
& =\lim _{x \rightarrow \infty} \frac{\sqrt{x}(x+c-x)}{\sqrt{x+c}+\sqrt{x}} \\
& =\lim _{x \rightarrow \infty} \frac{c \sqrt{x}}{\sqrt{x+c}+\sqrt{x}} \\
& =\lim _{x \rightarrow \infty} \frac{c}{\sqrt{1+\frac{c}{x}}+1} \\
& =\lim _{x \rightarrow \infty} \frac{c}{\sqrt{1+\frac{c}{x}}+1}=\frac{c}{2}
\end{aligned}
$
Question 2.
Evaluate: $\lim _{x \rightarrow \infty}\left(\sqrt{x^2+x+1}-\sqrt{x^2+1}\right)$

Solution:
Solution:
Here the expression assumes the form $\infty-\infty$ as $\mathrm{x} \rightarrow \infty$. So, we first reduce it to the rational form $\frac{f(x)}{g(x)}$
$
\begin{aligned}
& \lim _{x \rightarrow \infty} \sqrt{x^2+x+1}-\sqrt{x^2+1}=\lim _{x \rightarrow \infty} \frac{\left\{\sqrt{x^2+x+1}-\sqrt{x^2+1}\right\}}{\left\{\sqrt{x^2+x+1}+\sqrt{x^2+1}\right\}}\left\{\sqrt{x^2+x+1}+\sqrt{x^2+1}\right\} \\
& =\lim _{x \rightarrow \infty} \frac{x^2+x+1-x^2-1}{\sqrt{x^2+x+1}+\sqrt{x^2+1}} \\
& =\lim _{x \rightarrow \infty} \frac{x}{\sqrt{x^2+x+1}+\sqrt{x^2+1}}=\lim _{x \rightarrow \infty} \frac{x}{\sqrt[x]{1+\frac{1}{x}+\frac{1}{x^2}}+\sqrt{1+\frac{1}{x^2}}} \\
& =\lim _{x \rightarrow \infty} \frac{1}{\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+\sqrt{1+\frac{1}{x^2}}} \\
& =\frac{1}{1+1}=\frac{1}{2} \\
&
\end{aligned}
$

Question 3.
Evaluate $\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}$
Solution:
We have
$
\begin{aligned}
\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2} & =\lim _{n \rightarrow \infty} \frac{1}{n^2} \times \frac{n(n+1)}{2} \quad\left[\because 1+2+\ldots+n=\frac{n(n+1)}{2}\right] \\
& =\lim _{n \rightarrow \infty} \frac{1}{2}\left(1+\frac{1}{n}\right)=\frac{1}{2}
\end{aligned}
$
Question 4.
Evaluate:
$
\lim _{n \rightarrow \infty} \frac{n !}{(n+1) !-n !}
$
Solution:
We have,
$
\begin{aligned}
\lim _{n \rightarrow \infty} \frac{n !}{(n+1) !-n !} & =\lim _{n \rightarrow \infty} \frac{n !}{(n+1) n !-n !}=\lim _{n \rightarrow \infty} \frac{1}{n+1-1} \\
& =\lim _{n \rightarrow \infty} \frac{1}{n}=0
\end{aligned}
$
Question 5.
Evaluate: $\lim _{x \rightarrow-\infty}\left(\sqrt{x^2-x+1}+x\right)$

Solution:
$
\begin{aligned}
\lim _{x \rightarrow-\infty}\left(\sqrt{x^2-x+1}+x\right) & =\lim _{x \rightarrow-\infty} \frac{\left\{\sqrt{x^2-x+1}+x\right\}\left\{\sqrt{x^2-x+1}-x\right\}}{\left\{\sqrt{x^2-x+1}-x\right\}} \\
& =\lim _{x \rightarrow-\infty} \frac{x^2-x+1-x^2}{\sqrt{x^2-x+1}-x} \\
& =\lim _{x \rightarrow-\infty} \frac{-x+1}{\sqrt{x^2-x+1}-x} \\
& =\lim _{x \rightarrow-\infty} \frac{-\frac{x}{|x|}+\frac{1}{|x|}}{\sqrt{x^2-x+1}-\frac{x}{|x|}}
\end{aligned}
$
(Dividing numerator and denominator by $|x|$ )
$
\begin{aligned}
& =\lim _{x \rightarrow-\infty} \frac{\frac{-x}{-x}-\frac{1}{x}}{\sqrt{\frac{x^2}{x^2}-\frac{x}{x^2}+\frac{1}{x^2}+\frac{x}{x}}}\left(\therefore \sqrt{x^2}=|x|=-x \text { for } x<0\right) \\
& =\lim _{x \rightarrow-\infty} \frac{1-\frac{1}{x}}{\sqrt{1-\frac{1}{x}}+\frac{1}{x^2}+1}=\frac{1}{2} \\
&
\end{aligned}
$
Question 6.
Evaluate: $\lim _{x \rightarrow \infty} \frac{(x+1)^{10}+(x+2)^{10}+\ldots+(x+100)^{10}}{x^{10}+10^{10}}$
Solution:

$\begin{aligned}
& \lim _{x \rightarrow \infty} \frac{(x+1)^{10}+(x+2)^{10}+\ldots+(x+100)^{10}}{x^{10}+10^{10}} \\
&=\left.\lim _{x \rightarrow \infty} \frac{x^{10}\left[\left(1+\frac{1}{x}\right)^{10}+\left(1+\frac{2}{x}\right)^{10}+\ldots .\left(1+\frac{100}{x}\right)^{10}\right]}{x^{10}\left(1+\frac{10^{10}}{x^{10}}\right)}\right] \\
&= \lim _{x \rightarrow \infty} \frac{\left(1+\frac{1}{x}\right)^{10}+\left(1+\frac{2}{x}\right)^{10} \ldots .+\left(1+\frac{100}{x}\right)^{10}}{\left(1+\left(\frac{10}{x}\right)^{10}\right)} \\
&= \frac{1+1+\ldots+1(100 \text { times })}{1+0}=\frac{100}{1}=100
\end{aligned}$