Question

Question!
Evaluate the following
$\sf \lim_{h\to 0}\dfrac{(a+h)^2 \;\sin(a+h)-a^2 \sin a}{h}$

Answer 1

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

Given expression is

$\rm :\longmapsto\:\displaystyle\lim _{h \to 0}\dfrac{(a+h)^2 \;\sin(a+h)-a^2 \sin a}{h}$

If we substitute directly h = 0, we get

$\rm \:  =  \:\dfrac{(a+0)^2 \;\sin(a+0)-a^2 \sin a}{0}$

$\rm \:  =  \:\dfrac{a^2 \;sina-a^2 \sin a}{0}$

$\rm \:  =  \:\dfrac{0}{0}$

which is indeterminant form.

So,

$\rm :\longmapsto\:\displaystyle\lim _{h \to 0}\dfrac{(a+h)^2 \;\sin(a+h)-a^2 \sin a}{h}$

can be rewritten as

$\rm =\displaystyle\lim _{h \to 0}\dfrac{( {a}^{2}+{h}^{2}+2ah)\sin(a+h)-a^2 \sin a}{h}$

$\rm =\displaystyle\lim _{h \to 0}\dfrac{{a}^{2}\sin(a+h) + {h}^{2}sin(a + h) + 2ahsin(a + h) -a^2 \sin a}{h}$

can be re-arranged as

$\rm =\displaystyle\lim _{h \to 0}\dfrac{{a}^{2}(\sin(a+h) - sina) + {h}^{2}sin(a + h) + 2ahsin(a + h)}{h}$

$\rm ={a}^{2}\displaystyle\lim _{h \to 0} \frac{sin(a + h) - sina}{h} + \displaystyle\lim _{h \to 0} \frac{ {h}^{2}sin(a + h) }{h} + \displaystyle\lim _{h \to 0} \frac{2asin(a + h)}{h}$

We know,

$\boxed{ \tt{ \: sinx - siny = 2cos\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg]}}$

So, using this identity, we get

$\rm = {a}^{2}\displaystyle\lim _{h \to 0} \frac{2cos\bigg[\dfrac{a + h + a}{2} \bigg]sin\bigg[\dfrac{a + h - a}{2} \bigg]}{h} + \displaystyle\lim _{h \to 0}hsin(a + h) + 2asina$

$\rm = {a}^{2}\displaystyle\lim _{h \to 0} \frac{2cos\bigg[\dfrac{2a + h}{2} \bigg]sin\bigg[\dfrac{h}{2} \bigg]}{h} + 0 + 2asina$

$\rm \:  =  \:2 {a}^{2}\displaystyle\lim _{h \to 0}cos\bigg[\dfrac{2a + h}{2} \bigg] \times \displaystyle\lim _{h \to 0} \frac{sin\bigg[\dfrac{h}{2} \bigg]}{ \dfrac{h}{2} \times 2} + 2asina$

We know

$\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim _{x \to 0} \: \frac{sinx}{x} \: = \: 1 \: }}$

So, using this, we get

$\rm \:  =  \: 2{a}^{2}cos\bigg[\dfrac{2a + 0}{2} \bigg] \times \dfrac{1}{2} + 2asina$

$\rm \:  =  \: {a}^{2} \: cosa + 2a \: sina$

Hence,

$\boxed{ \tt{ \: \displaystyle\lim _{h \to 0}\dfrac{(a+h)^2 \;\sin(a+h)-a^2 \sin a}{h} = {a}^{2}cosa + 2asina}}$

Additional Information :-

$\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim _{x \to 0} \: \frac{tanx}{x} \: = \: 1 \: }}$

$\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim _{x \to 0} \: \frac{log(1 + x)}{x} \: = \: 1 \: }}$

$\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim _{x \to 0} \: \frac{ {e}^{x} - 1}{x} \: = \: 1 \: }}$

$\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim _{x \to 0} \: \frac{ {a}^{x} - 1}{x} \: = \: loga \: }}$

$\boxed{ \tt{ \: sinx + siny = 2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}}$

$\boxed{ \tt{ \: cosx + cosy = 2cos\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}}$

$\boxed{ \tt{ \: cosx - cosy = - \: 2sin\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg]}}$