Question

Please solve this problem
Evaluate the following limits
lim x->0 x/sin^ -1 x​

Answer 1

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

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{x}{ {sin}^{ - 1}x}$

To evaluate this limit, we use Method of Substitution

So, Substitute

$\red{\rm :\longmapsto\: {sin}^{ - 1}x = y}$

$\red{\rm :\longmapsto\: x = {sin}y}$

$\red{\rm :\longmapsto\: As \: x \: \to \: 0, \: so \: y \: \to \: 0}$

So, above expression can be rewritten as

$\rm \:  =  \: \displaystyle\lim_{y \to 0}\rm \frac{siny}{ {sin}^{ - 1}(siny) }$

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

$\rm \:  =  \: 1$

Hence,

$\purple{\rm\implies \:\boxed{\tt{ \: \: \: \: \: \displaystyle\lim_{x \to 0}\rm \frac{x}{ {sin}^{ - 1}x} = 1} \: \: \: \: \: \: }}$

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More to know

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

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

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

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

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

Answer 2

Step-by-step explanation:

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

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{x}{ {sin}^{ - 1}x}$

To evaluate this limit, we use Method of Substitution

So, Substitute

$\blue{\rm :\longmapsto\: {sin}^{ - 1}x = y}$

$\blue{\rm :\longmapsto\: x = {sin}y}$

$\blue{\rm :\longmapsto\: As \: x \: \to \: 0, \: so \: y \: \to \: 0}$

So, above expression can be rewritten as

$\rm \:  =  \: \displaystyle\lim_{y \to 0}\rm \frac{siny}{ {sin}^{ - 1}(siny) }$

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

$\rm \:  =  \: 1$

Hence,

$\green{\rm\implies \:\boxed{\tt{ \: \: \: \: \: \displaystyle\lim_{x \to 0}\rm \frac{x}{ {sin}^{ - 1}x} = 1} \: \: \: \: \: \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More to know

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

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

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

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

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