Exercise 7.5-Additional Problems - Chapter 7 - Applications of Differential Calculus - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Problems
Question 1.
$
\lim _{x \rightarrow \pi / 2} \frac{\log (\sin x)}{(\pi-2 x)^2}
$
Solution:

$
\begin{aligned}
& \text { It is of the form } \frac{0}{0} \\
& \qquad \begin{aligned}
\lim _{x \rightarrow \pi / 2} \frac{\log (\sin x)}{(\pi-2 x)^2} & =\lim _{x \rightarrow \pi / 2} \frac{\frac{1}{\sin x} \cos x}{2(\pi-2 x) \times(-2)} \\
& =\lim _{x \rightarrow \pi / 2} \frac{\cot x}{-4(\pi-2 x)}=\frac{0}{0} \\
& =\lim _{x \rightarrow \pi / 2} \frac{-\operatorname{cosec}^2 x}{-4 \times-2}=\frac{-1}{8}
\end{aligned}
\end{aligned}
$
Note that here l'Hopital's rule, applied yields the result
Question 2.
Evaluate: $\lim _{x \rightarrow 0}(\cot x)^{\sin x}$
Solution:
$\lim _{x \rightarrow 0}(\cot x)^{\sin x}$ is of the type $\infty^0$.
Let $y=(\cot x)^{\sin x} \Rightarrow \log y=\sin x \log (\cot x)$
$
\begin{aligned}
\lim _{x \rightarrow 0}(\log y) & =\lim _{x \rightarrow 0} \sin x \log (\cot x) \\
& =\lim _{x \rightarrow 0} \frac{\log (\cot x)}{\operatorname{cosec} x} \text { is of the type } \frac{\infty}{\infty}
\end{aligned}
$
Applying $l$ Hôpital's rule.
$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\log (\cot x)}{\operatorname{cosec} x} & =\lim _{x \rightarrow 0} \frac{\frac{1}{\cot x}\left(-\operatorname{cosec}^2 x\right)}{-\operatorname{cosec} x \cot x} \\
& =\lim _{x \rightarrow 0} \frac{\sin x}{\cos x} \times \frac{1}{\cos x}=\frac{0}{1}=0
\end{aligned}
$
(i.e.,) $\lim _{x \rightarrow 0} \log y=0$
By Composite Function Theorem, we have
$
0=\lim _{x \rightarrow 0} \log y=\log \left(\lim _{x \rightarrow 0} y\right) \Rightarrow \lim _{x \rightarrow 0} y=e^0=1
$

Question 3.
$
\lim _{x \rightarrow \infty} \frac{\log e^x}{x}
$
Solution:
$
\lim _{x \rightarrow \infty} \frac{\log e^x}{x}=\frac{\log \infty}{\infty}=\frac{\infty}{\infty} ; \text { So, } \lim _{x \rightarrow \infty} \frac{\frac{1}{x}}{1}=\frac{1}{\infty}=0
$
Question 4.
$
\lim _{x \rightarrow 0^{+}} x^2 \log e^x
$
Solution:
$
\lim _{x \rightarrow 0^{+}} x^2 \log x=\text { (0) } \log 0
$
$=0 \times(-\infty)$ which is indeterminate form.
So,
$
\begin{aligned}
\lim _{x \rightarrow 0^{+}} x^2 \log _e x & =\lim _{x \rightarrow 0^{+}} \frac{\log e^x}{\frac{1}{x^2}} \\
= & \lim _{x \rightarrow 0^{+}} \frac{\frac{1}{x}}{\frac{-2}{x^3}}=\frac{1 \times x^3}{x \times(-2)}=\frac{x^2}{-2}=\frac{0}{-2}=0
\end{aligned}
$

Question 5.
$
\lim _{x \rightarrow 1} x^{\frac{1}{x-1}}
$
Solution:
Let $y=\lim _{x \rightarrow 1} x^{\frac{1}{x-1}}$
Taking log on both sides,
$
\log y=\lim _{x \rightarrow 1} \log x^{\frac{1}{x-1}}=\frac{1}{x-1} \log x=\frac{\log x}{x-1}=\frac{\log 1}{1-1}=\frac{0}{0}
$
So, applying L.H. rule
$
\lim _{x \rightarrow 1} \frac{\frac{1}{x}}{1}=\frac{1}{1}=1 ; \text { i.e., } \log y=1 ; \therefore y=e^1=e
$

Question 6.
$
\lim _{x \rightarrow 0}(\cos x)^{\frac{1}{x}}
$
Solution:
Let $y=\lim _{x \rightarrow 0}(\cos x)^{\frac{1}{x}}$
Taking log on both sides,
$
\begin{aligned}
\log y & =\lim _{x \rightarrow 0} \log (\cos x)^{\frac{1}{x}}=\lim _{x \rightarrow 0} \frac{1}{x} \log (\cos x) \\
& =\lim _{x \rightarrow 0} \frac{\log (\cos x)}{x}=\frac{\log (1)}{0}=\frac{0}{0}
\end{aligned}
$
Applying L.H. rule,
$
=\lim _{x \rightarrow 0} \frac{\frac{1}{\cos x}(-\sin x)}{1} \cdot=\lim _{x \rightarrow 0} \frac{-\sin x}{\cos x}=0
$
So, $\log y=0 \Rightarrow y=e^0=1$