Exercise 7.5 - Chapter 7 Applications of Differential Calculus 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Applications ofDifferential Calculus
$
\text { Ex } 7.5
$

Question 1.
Evaluate the following limits, if necessary use l'Hopital Rule.
$
\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2}
$
Solution:
$\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2}=\frac{1-1}{0}=\frac{0}{0}=$ indeterminate form
so (Applying l'Hôpital Rule we get)
$
\lim _{x \rightarrow 0} \frac{\sin x}{2 x}=\frac{0}{0}=\text { indeterminate form }
$
so l'Hôpital Rule we get
$
\lim _{x \rightarrow 0}=\frac{\cos x}{2}=\frac{1}{2}
$

Question 2.
$
\lim _{x \rightarrow \infty} \frac{2 x^2-3}{x^2-5 x+3}
$
Solution:
$
\text { Put } \begin{aligned}
x=\frac{1}{y} \text { so } x \rightarrow \infty & \Rightarrow y \rightarrow 0 \\
& =\lim _{y \rightarrow 0} \frac{\frac{2}{y^2}-3}{\frac{1}{y^2}-\frac{5}{y}+3} \\
& =\lim _{y \rightarrow 0} \frac{\frac{1}{y^2}\left(2-3 y^2\right)}{\frac{1}{y^2}\left(1-5 y+3 y^2\right)}=\frac{2}{1}=2
\end{aligned}
$

Question 3.
$
\lim _{x \rightarrow \infty} \frac{x}{\log x}
$
Solution:
$
\lim _{x \rightarrow \infty} \frac{x}{\log x}=\left(\frac{\infty}{\infty}\right)=\text { indeterminate form }
$
$\therefore$ Applying L.H. Rule
$
\operatorname{Lt}_{x \rightarrow \infty} \frac{1}{1 / x}=\lim _{x \rightarrow \infty} x=\infty
$

Question 4.
$
\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sec x}{\tan x}
$
Solution:
$
\begin{aligned}
\lim _{x \rightarrow \frac{\pi}{2}} \frac{1}{\cos x} \times \frac{\cos x}{\sin x} & =\lim _{x \rightarrow \frac{\pi}{2}} \frac{1}{\sin x} \\
& =\frac{1}{1}=1
\end{aligned}
$
Question 5.
$
\lim _{x \rightarrow \infty} e^{-x} \sqrt{x}
$
Solution:
$
\lim _{x \rightarrow \infty} e^{-x} \sqrt{x}=\frac{\sqrt{x}}{e^x}=\left(\frac{\infty}{\infty}\right)=\text { indeterminate form }
$
Applying L.H. Rule
$
\begin{aligned}
& \lim _{x \rightarrow \infty} \frac{\frac{1}{2 \sqrt{x}}}{e^x}=\lim _{x \rightarrow \infty} \frac{1}{2 \sqrt{x} e^x} \\
& \text { as } x \rightarrow \infty \text {, denominator } \rightarrow 0
\end{aligned}
$

Question 6.
$
\lim _{x \rightarrow 0}\left(\frac{1}{\sin x}-\frac{1}{x}\right) .
$
Solution:
$\lim _{x \rightarrow 0} \frac{x-\sin x}{x \sin x}=\left(\frac{0}{0}\right)=$ indeterminate form
Applying L.H. Rule
$\lim _{x \rightarrow 0} \frac{1-\cos x}{x \cos x+\sin x}=\frac{1-1}{0+0}=\frac{0}{0}$ indeterminate form
Again applying L.H. Rule
$
\frac{\sin x}{-x \sin x+\cos x+\cos x}=\frac{0}{2}=0
$
Question 7.
$
\lim _{x \rightarrow 1^{+}}\left(\frac{2}{x^2-1}-\frac{x}{x-1}\right)
$
Solution:
$
\lim _{x \rightarrow 1^{+}} \frac{2}{x^2-1}-\lim _{x \rightarrow 1^{+}} \frac{x}{x-1}
$
Applying L.H. Rule
$
\lim _{x \rightarrow 1^{+}} \frac{0}{2 x}-\lim _{x \rightarrow 1^{+}} \frac{1}{1}=0-1=-1
$

Question 8.
$
\lim _{x \rightarrow 0^{+}} x^x \text {. }
$
Solution:
Let $y=x^x$
Taking log on both sides
$
\begin{aligned}
& \log y=\log x^x=x \log x \\
& \lim _{x \rightarrow 0^{+}} \log y=\lim _{x \rightarrow 0^{+}} \frac{\log x}{\frac{1}{x}}=\frac{-\infty}{\infty} \\
& \text { Applying L.H. Rule } \\
& \lim _{x \rightarrow 0^{+}} \frac{1 / x}{-1 / x^2}=\lim _{x \rightarrow 0^{+}} \frac{1}{x}\left(\frac{-x^2}{1}\right) \\
& =\lim _{x \rightarrow 0^{+}}(-x)=0 \\
& \log y=0 \Rightarrow y=e^0=1 \\
&
\end{aligned}
$
Question 9.
$
\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^x
$

Solution:
$
\text { Let } y=\left(1+\frac{1}{x}\right)^x
$

Taking log on both sides
$
\log y=\log \left(1+\frac{1}{x}\right)^x=x \log \left(1+\frac{1}{x}\right)
$
$
\begin{aligned}
& \lim _{x \rightarrow \infty} \log y=\lim _{x \rightarrow \infty} \frac{\log (1+1 / x)}{1 / x} \\
& \text { Applying L.H. Rule } \Rightarrow \frac{\lim _{x \rightarrow \infty}\left(\frac{1}{1+1 / x}\right)\left(\frac{-1}{x^2}\right)}{\left(\frac{-1}{x^2}\right)} \\
& \begin{array}{l}
\lim _{x \rightarrow \infty} \log y=\lim _{x \rightarrow \infty}\left(\frac{1}{1+1 / x}\right)=1 \\
y=e^1=e
\end{array} \\
& \log y=1 \Rightarrow y=e^1=e \\
&
\end{aligned}
$

Question 10.
$
\lim _{x \rightarrow \frac{\pi}{2}}(\sin x)^{\tan x}
$
Solution:
$
\text { Let } y=(\sin x)^{\tan x}
$
Taking log on both sides
$
\begin{aligned}
\log y & =\log (\sin x)^{\tan x}=\tan x \log (\sin x) \\
& =\frac{\log \sin x}{\cot x} \\
\lim _{x \rightarrow \frac{\pi}{2}} \log y & =\lim _{x \rightarrow \frac{\pi}{2}} \frac{\log \sin x}{\cot x}=\frac{0}{0}=\text { indeterminate form }
\end{aligned}
$
Applying L.H. Rule
$
\begin{gathered}
\lim _{x \rightarrow \frac{\pi}{2}} \log y=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\frac{1}{\sin x} \cos x}{-\operatorname{cosec}^2 x}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{-\operatorname{cosec} x}=0 \\
\text { (i.e.) } \log y=0 \Rightarrow y=e^0=1
\end{gathered}
$

Question 11.
$
\lim _{x \rightarrow 0^{+}}(\cos x)^{\frac{1}{x^2}}
$

Solution:
$
\lim _{x \rightarrow 0^{+}}(\cos x) x^2 \text {. }
$
Let $y=(\cos x)^{\frac{1}{x^2}}$
Taking log on both sides
$
\begin{aligned}
\log y & =\log (\cos x)^{\frac{1}{x^2}}=\frac{1}{x^2} \log \cos x \\
\log y & =\frac{\log (\cos x)}{x^2} \\
\lim _{x \rightarrow 0^{+}} \log y & =\lim _{x \rightarrow 0^{+}} \frac{\log (\cos x)}{x^2}=\frac{0}{0}=\text { indeterminate form }
\end{aligned}
$
So Applying L.H. Rule
$
\lim _{x \rightarrow 0^{+}} \log y=\lim _{x \rightarrow 0^{+}} \frac{\frac{1}{\cos x}(-\sin x)}{2 x}=\frac{-\tan x}{2 x}=\frac{0}{0}=\text { indeterminate form }
$
Again Applying L.H. Rule
$
\begin{aligned}
\lim _{x \rightarrow 0^{+}} \log y & =\lim _{x \rightarrow 0^{+}} \frac{-\sec ^2 x}{2}=\frac{-1}{2} \\
\log y & =-\frac{1}{2} \\
\Rightarrow y & =e^{-\frac{1}{2}}=\frac{1}{e^{1 / 2}}=\frac{1}{\sqrt{e}}
\end{aligned}
$

Question 12.
If an initial amount $\mathrm{A}_0$ of money is invested at an interest rate $\mathrm{r}$ compounded $\mathrm{n}$ times a year, the value
of the investment after t years is $
\mathrm{A}=\mathrm{A}_0\left(1+\frac{r}{n}\right)^{n t}
$ If the interest is compounded continuously, (that is as $\mathrm{n} \rightarrow \infty$ ), show that the amount after t years is $\mathrm{A}=\mathrm{A}_0 \mathrm{e}^{\mathrm{rt}}$.
Solution:

$
\begin{aligned}
\text { Given } \mathrm{A} & =\lim _{n \rightarrow \infty} \mathrm{A}_0\left(1+\frac{r}{n}\right)^{n t} \\
\text { Let } y & =\mathrm{A}_0\left(1+\frac{r}{n}\right)^{n t} \\
\therefore \log y & =\log \mathrm{A}_0+(n t) \log \left(1+\frac{r}{n}\right) \\
\lim _{n \rightarrow \infty} \log y & =\log \mathrm{A}_0+\lim _{n \rightarrow \infty} \log \left(\frac{1+\frac{r}{n}}{1 / n t}\right)
\end{aligned}
$
Applying L.H. Rule

$\begin{aligned}
& \log y=\log \mathrm{A}_0+\lim _{n \rightarrow \infty} \frac{\frac{1}{\left(1+\frac{r}{n}\right)}\left(0-\frac{r}{n^2}\right)}{\frac{-1}{t^2}} \\
& \text { (i.e.,) } \lim _{n \rightarrow \infty} \log y=\log \mathrm{A}_0+\lim _{n \rightarrow \infty}\left(\frac{1}{1+\frac{r}{n}}\right)\left(\frac{-r}{n^2}\right)\left(-\operatorname{tn}^2\right) \\
& \begin{aligned}
\text { (i.e.,) } & \lim _{n \rightarrow \infty} \log y=\log \mathrm{A}_0+r t \lim _{n \rightarrow \infty}\left(\frac{1}{1+\frac{r}{n}}\right) \\
\text { (i.e.,) } & \lim _{n \rightarrow \infty} \log y=\log \mathrm{A}_0+r t(1)
\end{aligned} \\
& \log \left(\lim _{n \rightarrow \infty} y\right)=\log \mathrm{A}_0+r t \\
& =\lim _{n \rightarrow \infty} \mathrm{A}_0\left(1+\frac{r}{n}\right)^{n t}=e^{\log \mathrm{A} 0+r t} \\
& =e^{\log \mathrm{A} 0} \cdot e^{r t} \\
& \text { (i.e.,) } \\
& \mathrm{A}=\mathrm{A}_0 e^{r t} \\
&
\end{aligned}$