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$\sf \: \lim _{x \longrightarrow - \infty }( \sin( \frac{1}{ \theta} ) )$
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Answer 1

Appropriate Question :-

Evaluate the following :-

$\rm :\longmapsto\:\displaystyle\lim_{x \to - \infty } \: sin \: \frac{1}{x}$

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

$\rm :\longmapsto\:\displaystyle\lim_{x \to - \infty } \: sin \: \frac{1}{x}$

To evaluate this limit, we use Method of Substitution

So, Substitute

$\red{\sf :\longmapsto\:\dfrac{1}{x} \: = \: - \: y \: \: as \: x \to \: - \infty , \: \: so \: y \: \to \: 0}$

So, above expression can be rewritten as

$\rm \: = \: \displaystyle\lim_{y \to 0 } \: sin( - y)$

$\rm \: = \: - \: \displaystyle\lim_{y \to 0 } \: siny$

$\rm \: = \: - \: sin0$

$\rm \: = \: 0$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to - \infty } \bf \: \: sin \: \frac{1}{x} = 0 \: }}$

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Additional Information :-

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

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

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

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

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

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