Question

Prove that
Lt/(h->0): [(a+h)^2 sin(a+h)-a^2 sina]/h
= 2a sina + a^2 cosa
​

Answer 1

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

Consider,

$\red{\rm :\longmapsto\:\displaystyle\lim_{h \to 0}\sf \frac{ {(a + h)}^{2} sin(a + h) - {a}^{2}sina}{h} \: }$

If we substitute directly h = 0, we get

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

$\red{\rm \: = \: \sf \dfrac{ {a}^{2} sina - {a}^{2}sina}{0} \: }$

$\red{\rm \: = \: \sf \dfrac{ 0}{0} \: }$

which is indeterminant form.

So, Consider again

$\red{\rm :\longmapsto\:\displaystyle\lim_{h \to 0}\sf \frac{ {(a + h)}^{2} sin(a + h) - {a}^{2}sina}{h} \: }$

can be rewritten as

$=\displaystyle\lim_{h \to 0}\sf \frac{({a}^{2}+{h}^{2}+2ah)sin(a + h)-{a}^{2}sina}{h}$

$\sf=\displaystyle\lim_{h \to 0}\sf \frac{{a}^{2}sin(a + h)+{h}^{2}sin(a + h)+2ahsin(a + h)-{a}^{2}sina}{h}$

can be rewritten as

$\sf=\displaystyle\lim_{h \to 0}\sf \frac{{a}^{2}sin(a + h)-{a}^{2}sina+{h}^{2}sin(a + h)+2ahsin(a + h)}{h}$

$\sf=\displaystyle\lim_{h \to 0}\sf \frac{{a}^{2}sin(a + h)-{a}^{2}sina}{h} + \displaystyle\lim_{h \to 0}\sf \frac{{h}^{2}sin(a + h)+2ahsin(a + h)}{h}$

$\sf=\displaystyle\lim_{h \to 0}\sf \frac{{a}^{2}[sin(a + h)-sina]}{h} + \displaystyle\lim_{h \to 0}\sf \frac{h[hsin(a + h)+2asin(a + h)]}{h}$

$\sf= {a}^{2} \displaystyle\lim_{h \to 0}\sf \frac{sin(a + h)-sina}{h} + \displaystyle\lim_{h \to 0}\sf \frac{[hsin(a + h)+2asin(a + h)]}{1}$

We know

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

$\sf= {a}^{2} \displaystyle\lim_{h \to 0}\sf \frac{2cos\bigg[\dfrac{a + h + a}{2} \bigg]sin\bigg[\dfrac{a + h - a}{2} \bigg]}{h} +2asina$

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

$\sf= 2{a}^{2}cosa \displaystyle\lim_{h \to 0}\sf \frac{sin\bigg[\dfrac{h}{2} \bigg]}{\dfrac{h}{2} } \times \dfrac{1}{2} +2asina$

We know,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 0}\sf \frac{sinx}{x} = 1 \: }}}$

So, using this, we get

$\rm \: = \sf {a}^{2} cosa \times 1 + 2a \: sina$

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

Hence,

$\red{\rm \:\boxed{ \tt{ \: \displaystyle\lim_{h \to 0}\sf \frac{ {(a + h)}^{2} sin(a + h) - {a}^{2}sina}{h} = {a}^{2}cosa + 2a \: sina \: }}}$

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Additional Information :-

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 0}\sf \frac{sinx}{x} = 1 \: }}}$

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 0}\sf \frac{tanx}{x} = 1 \: }}}$

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

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 0}\sf \frac{ {e}^{x} - 1}{x} = 1 \: }}}$

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

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 0}\sf \frac{sin ^{ - 1} x}{x} = 1 \: }}}$

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 0}\sf \frac{tan ^{ - 1} x}{x} = 1 \: }}}$