Test Index

UPSC NDA Math Model Paper 3

© examsnet.com

Question : 88 of 120

Marks: +1, -0

Find the value of

\lim\limits_{h\to 0}\left\{\frac{1}{h\sqrt[3]{8+h}}-\frac{1}{2h}\right\}

$\frac{-1}{24}$
$\frac{-1}{48}$
$\frac{-1}{12}$
$\frac{-1}{8}$

Solution:

Let

L=\lim\limits_{h\to 0}\left\{\frac{1}{h\sqrt[3]{8+h}}-\frac{1}{2h}\right\}

=\lim\limits_{h\to 0}\frac{1}{h}\times\left[\frac{1}{(8+h)^{1/3}}-\frac{1}{2}\right]

=\lim\limits_{h\to 0}\frac{1}{h}\times\left[\frac{2-(8+h)^{1/3}}{2(8+h)^{1/3}}\right]

=\lim\limits_{h\to 0}\frac{1}{h}\times\left[\frac{2-2\left(1+\frac{h}{8}\right)^{1/3}}{2(8+h)^{1/3}}\right]

=\lim\limits_{h\to 0}\frac{1}{h}\times\left[\frac{1-\left(1+\frac{h}{8}\right)^{1/3}}{(8+h)^{1/3}}\right]

As we know

(1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+\cdots

Therefore,

\left(1+\frac{h}{8}\right)^{1/3}=1+\frac{1}{3}\cdot\frac{h}{8}+\frac{\frac{1}{3}\left(\frac{1}{3}-1\right)}{2}

\cdot\left(\frac{h}{8}\right)^2+\cdots

\left(1+\frac{h}{8}\right)^{1/3}=1+\frac{h}{24}-\frac{1}{9}\cdot\frac{h^2}{64}

+\cdots

L=\lim\limits_{h\to 0}\frac{1}{h}\times\left[\frac{1-\left(1+\frac{h}{24}-\frac{1}{9}\cdot\frac{h^2}{64}+\cdots\right)}{(8+h)^{1/3}}\right]

=\lim\limits_{h\to 0}\frac{1}{h}\times\left[\frac{-\frac{h}{24}+\frac{1}{9}\cdot\frac{h^2}{64}+\cdots}{(8+h)^{1/3}}\right]

=\lim\limits_{h\to 0}\frac{1}{h}\times h\times\left[\frac{-\frac{1}{24}+\frac{1}{9}\cdot\frac{h}{64}+\cdots}{(8+h)^{1/3}}\right]

=\lim\limits_{h\to 0}\left[\frac{-\frac{1}{24}+\frac{1}{9}\cdot\frac{h}{64}+\cdots}{(8+h)^{1/3}}\right]

=\left[\frac{-\frac{1}{24}+\frac{1}{9}\cdot\frac{0}{64}+0+\cdots}{(8+0)^{1/3}}\right]

=\frac{-\frac{1}{24}}{2}=\frac{-1}{48}

© examsnet.com

Go to Question:

Prev Question

Next Question