Question

lim x->0 (1-cosax)/xsin2x

Answer 1

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

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{1 - cos \: ax}{x \: sin2x} \: }$

If we substitute directly x = 0, we get

$\rm \: = \: \dfrac{1 - cos0}{0 \times sin0}$

$\rm \: = \: \dfrac{1 - 1}{0}$

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

which is indeterminant form.

So, Consider again

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{1 - cos \: ax}{x \: sin2x} \: }$

We know,

$\boxed{ \tt{ \: 1 - cos2x = {2sin}^{2}x \: }}$

So, using this identity, we get

$\rm \: = \: \dfrac{ {2sin}^{2} \dfrac{ax}{2} }{x \times \red{\dfrac{sin2x}{2x}} \times 2x}$

We know,

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

So, using this identity, we get

$\rm \: = \: \displaystyle\lim_{x \to 0}\rm \frac{sin \dfrac{ax}{2} \times sin \dfrac{ax}{2} }{ {x}^{2} }$

$\rm \: = \: \displaystyle\lim_{x \to 0}\rm \frac{sin \dfrac{ax}{2} \times sin \dfrac{ax}{2} }{ x \times x}$

$\rm \: = \: \displaystyle\lim_{x \to 0}\rm \frac{sin \dfrac{ax}{2} \times sin \dfrac{ax}{2} }{ \dfrac{ax}{2} \times \dfrac{ax}{2} } \times \dfrac{a}{2} \times \dfrac{a}{2}$

$\rm \: = \: \dfrac{ {a}^{2} }{4} \times \displaystyle\lim_{x \to 0}\rm \frac{sin\dfrac{ax}{2} }{\dfrac{ax}{2} } \times \displaystyle\lim_{x \to 0}\rm \frac{sin\dfrac{ax}{2} }{\dfrac{ax}{2} }$

$\rm \: = \: \dfrac{ {a}^{2} }{4} \times 1 \times 1$

$\rm \: = \: \dfrac{ {a}^{2} }{4}$

Hence,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 0}\rm \frac{1 - cos \: ax}{x \: sin2x} \: } = \frac{ {a}^{2} }{4} \: }}$

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More to Know :-

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

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

$\boxed{ \tt{ \: \displaystyle\lim_{x \: \longmapsto \: 0}\rm \: \frac{1 - cosx}{x} \: = \: 0 \: }}$

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

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

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