Question

Differential calculus: limits

$\rm\lim\limits_{h\to \infty} \left\{\dfrac{1}{h^3\sqrt{8+h}}-\dfrac{1}{2h}\right\}$

Answer 1

Appropriate Question :-

Evaluate the following :-

$\rm\lim\limits_{h\to 0} \left\{\dfrac{1}{h \: \sqrt[3]{8 + h} }-\dfrac{1}{2h}\right\}$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm\lim\limits_{h\to 0} \left\{\dfrac{1}{h \: \sqrt[3]{8 + h} }-\dfrac{1}{2h}\right\}$

can be further rewritten as

$\rm \: = \:\rm\lim\limits_{h\to 0} \left\{\dfrac{2 - \sqrt[3]{8 + h} }{2 \: h \: \sqrt[3]{8 + h} }\right\}$

$\rm \: = \:\dfrac{1}{2 \times \sqrt[3]{8} } \: \rm\lim\limits_{h\to 0} \left\{\dfrac{2 - \sqrt[3]{8 + h} }{\: h \: }\right\}$

$\rm \: = \:\dfrac{1}{2 \times 2} \: \rm\lim\limits_{h\to 0} \left\{\dfrac{ \sqrt[3]{8} - \sqrt[3]{8 + h} }{\: h \: }\right\}$

We know,

$\boxed{\sf{  \:\rm \: x - y = \frac{ {x}^{3} - {y}^{3} }{ {x}^{2} + xy + {y}^{2} } \: \: }}$

So, using this result in numerator, we get

$\rm \: = \: \dfrac{1}{4} \:\lim\limits_{h\to 0} \dfrac{8 - (8 + h)}{h\bigg[ {( \sqrt[3]{8 + h} )}^{2} + \sqrt[3]{8 + h} \sqrt[3]{8} + {( \sqrt[3]{8} )}^{2} \bigg]}$

$\rm \: = \: \dfrac{1}{4} \:\lim\limits_{h\to 0} \dfrac{8 - 8 - h}{h\bigg[ {( \sqrt[3]{8 + h} )}^{2} + \sqrt[3]{8 + h} \sqrt[3]{8} + {( \sqrt[3]{8} )}^{2} \bigg]}$

$\rm \: = \: \dfrac{1}{4} \:\lim\limits_{h\to 0} \dfrac{- h}{h\bigg[ {( \sqrt[3]{8 + h} )}^{2} + \sqrt[3]{8 + h} \sqrt[3]{8} + {( \sqrt[3]{8} )}^{2} \bigg]}$

$\rm \: = \: - \: \dfrac{1}{4} \:\lim\limits_{h\to 0} \dfrac{1}{\bigg[ {( \sqrt[3]{8 + h} )}^{2} + \sqrt[3]{8 + h} \sqrt[3]{8} + {( \sqrt[3]{8} )}^{2} \bigg]}$

$\rm \: = \: - \: \dfrac{1}{4} \times \dfrac{1}{\bigg[ {( \sqrt[3]{8 + 0} )}^{2} + \sqrt[3]{8 + 0} \sqrt[3]{8} + 4 \bigg]}$

$\rm \: = \: - \: \dfrac{1}{4} \times \dfrac{1}{\bigg[ {( \sqrt[3]{8} )}^{2} + \sqrt[3]{8} \times 2 + 4 \bigg]}$

$\rm \: = \: - \: \dfrac{1}{4} \times \dfrac{1}{\bigg[ {( 2 )}^{2} + 2 \times 2 + 4 \bigg]}$

$\rm \: = \: - \: \dfrac{1}{4} \times \dfrac{1}{\bigg[ 4 + 4 + 4 \bigg]}$

$\rm \: = \: - \: \dfrac{1}{4} \times \dfrac{1}{12}$

$\rm \: = \: - \: \dfrac{1}{48}$

Hence,

$\\ \rm\implies \:\boxed{\sf{  \:\rm\lim\limits_{h\to 0} \left\{\dfrac{1}{h \: \sqrt[3]{8 + h} }-\dfrac{1}{2h}\right\} = - \frac{1}{48} \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\boxed{\sf{  \:\rm \: \lim\limits_{x\to 0} \frac{sinx}{x} \: = \: 1 \: \: }}$

$\boxed{\sf{  \:\rm \: \lim\limits_{x\to 0} \frac{tanx}{x} \: = \: 1 \: \: }}$

$\boxed{\sf{  \:\rm \: \lim\limits_{x\to 0} \frac{log(1 + x)}{x} \: = \: 1 \: \: }}$

$\boxed{\sf{  \:\rm \: \lim\limits_{x\to 0} \frac{ {e}^{x} - 1}{x} \: = \: 1 \: \: }}$

$\boxed{\sf{  \:\rm \: \lim\limits_{x\to 0} \frac{ {a}^{x} - 1}{x} \: = \: loga \: \: }}$