Math, asked by khushi15686, 1 month ago

Evaluate the following limits

lim \: x \to \: 0 \:  \:  \:  {(sinx)}^{x}

Answers

Answered by mathdude500
5

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

Given expression is

\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \:  {(sinx)}^{x}

Let assume that,

\rm :\longmapsto\:y = \displaystyle\lim_{x \to 0} \:  {(sinx)}^{x}

Taking log on both sides, we get

\rm :\longmapsto\:logy = \displaystyle\lim_{x \to 0} \: log {(sinx)}^{x}

We know,

\boxed{ \tt{ \: log {x}^{y} = y \: logx \: }}

So, using this identity,

\rm :\longmapsto\:logy = \displaystyle\lim_{x \to 0} \: x \: log(sinx)

can be rewritten as

\rm :\longmapsto\:logy = \displaystyle\lim_{x \to 0} \:  \frac{logsinx}{ \dfrac{1}{x} }

We know,

\boxed{ \tt{ \: \dfrac{d}{dx}logx =  \frac{1}{x}}}

and

\boxed{ \tt{ \: \dfrac{d}{dx} \frac{1}{ {x}^{n} } =  \frac{ - n}{ {x}^{n + 1} }}}

So, using these, we get

\rm :\longmapsto\:logy = \displaystyle\lim_{x \to 0} \frac{ \dfrac{1}{sinx} \dfrac{d}{dx}sinx}{ \dfrac{ - 1}{ {x}^{2} } }

\rm :\longmapsto\:logy = \displaystyle\lim_{x \to 0} \frac{ \dfrac{1}{sinx}  \times cosx}{ \dfrac{ - 1}{ {x}^{2} } }

\rm :\longmapsto\:logy = \displaystyle\lim_{x \to 0} \frac{cotx}{ \dfrac{ - 1}{ {x}^{2} } }

\rm :\longmapsto\:logy = \displaystyle\lim_{x \to 0} \:  {x}^{2}  \: cotx

\rm :\longmapsto\:logy = \displaystyle\lim_{x \to 0} \: \frac{ {x}^{2} }{tanx}

\rm :\longmapsto\:logy = \displaystyle\lim_{x \to 0} \: \frac{x }{tanx}  \times x

\rm :\longmapsto\:logy = 1 \times 0

\rm :\longmapsto\:logy = 0

\bf\implies \:y =  {e}^{0} = 1

Hence,

\bf\implies \:\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \:  {(sinx)}^{x}  = e \:  \: }}

More to know :-

\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\  \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf  -  \: sinx \\ \\ \sf tanx & \sf  {sec}^{2}x \\ \\ \sf cotx & \sf  -  {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf  -  \: cosecx \: cotx\\ \\ \sf  \sqrt{x}  & \sf  \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf  {e}^{x}  & \sf  {e}^{x}  \end{array}} \\ \end{gathered}

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