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Answer 1

Step-by-step explanation:

We have,

$\lim_{x \rarr0} \frac{ \sqrt{ \cos(x) } - \sqrt[3]{ \cos(x) } }{ \sin^{2} (x) }$

$= \lim_{x \rarr0} \frac{ \dfrac{d}{dx} \{\sqrt{ \cos(x) } - \sqrt[3]{ \cos(x) } \} }{ \dfrac{d}{dx} \{\sin^{2} (x) \} }$

$= \lim_{x \rarr0} \frac{ \dfrac{ - \sin(x) }{2\sqrt{ \cos(x) }} + \dfrac{ \sin(x) }{3 \sqrt[3]{ \cos ^{2} (x) } } }{ 2\sin (x) \cos(x) }$

$= \lim_{x \rarr0} \frac{ \dfrac{ - 1 }{2\sqrt{ \cos(x) }} + \dfrac{ 1 }{3 \sqrt[3]{ \cos ^{2} (x) } } }{ 2\cos(x) }$

$= \frac{ \dfrac{ - 1 }{2\sqrt{ \cos(0) }} + \dfrac{ 1 }{3 \sqrt[3]{ \cos ^{2} (0) } } }{ 2\cos(0) }$

$= \frac{ \dfrac{ - 1 }{2\sqrt{ 1}} + \dfrac{ 1 }{3 \sqrt[3]{1 } } }{ 2(1) }$

$= \frac{ \dfrac{ - 1 }{2} + \dfrac{ 1 }{3 } }{ 2 }$

$= \frac{ \dfrac{ - 3 + 2 }{6} }{ 2 }$

$= \frac{ \dfrac{ - 1 }{6} }{ 2 }$

$= - \frac{ 1 }{ 12 }$

Answer 2

$\large\underline{\sf{Given \:Question - }}$

Evaluate

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \frac{ \sqrt{cosx} - \sqrt[3]{cosx} }{{sin}^{2}x }$

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

Given function is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \frac{ \sqrt{cosx} - \sqrt[3]{cosx} }{{sin}^{2}x }$

If we put directly x = 0, we get

$\rm \: = \: \dfrac{ \sqrt{cos0} - \sqrt[3]{cos0} }{{sin}^{2}0 }$

$\rm \:  =  \:\dfrac{1 - 1}{0}$

$\rm \:  =  \:\dfrac{0}{0} \: which \: is \: meaningless$

So,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \frac{ \sqrt{cosx} - \sqrt[3]{cosx} }{{sin}^{2}x }$

can be rewritten as

$\rm \:  =  \:\displaystyle\lim_{x \to 0} \frac{ \sqrt[3]{cosx}\bigg[\dfrac{ \sqrt{cosx} }{ \sqrt[3]{cosx}} - 1\bigg]}{1 - {cos}^{2}x}$

$\rm \:  =  \:\displaystyle\lim_{x \to 0} \sqrt[3]{cosx} \times \displaystyle\lim_{x \to 0} \frac{ \sqrt[6]{cosx} - 1}{1 - {cos}^{2} x}$

$\rm \:  =  - \:1\times \displaystyle\lim_{x \to 0} \frac{ \sqrt[6]{cosx} - 1}{{cos}^{2} x - 1}$

Now, to evaluate this limit, we use method of Substitution.

Let assume that

$\red{\rm :\longmapsto\: \sqrt[6]{cosx} = y}$

$\red{\rm :\longmapsto\:cosx = {y}^{6}, \: \: as \: x \to \: 0, \: \: y \: \to \: 1}$

So, above can be rewritten as

$\rm \:  =  \: - \: \displaystyle\lim_{y \to 1} \: \frac{y - 1}{ {y}^{12} - 1}$

can be rewritten as

$\rm \:  =  \: - \: \displaystyle\lim_{y \to 1} \: \dfrac{1}{\dfrac{ {y}^{12} - 1}{y - 1} }$

can be further rewritten as

$\rm \:  =  \: - \: \displaystyle\lim_{y \to 1} \: \dfrac{1}{\dfrac{ {y}^{12} - {1}^{12} }{y - 1} }$

We know,

$\red{\boxed{ \sf\:\displaystyle\lim_{x \to a} \: \frac{ {x}^{n} - {a}^{n} }{x - a} = {na}^{n - 1} }}$

So, using this we get

$\rm \:  =  \: - \: 1 \times \dfrac{1}{12 \times {(1)}^{12 - 1} }$

$\rm \:  =  \: - \: 1 \times \dfrac{1}{12 \times {(1)}^{11} }$

$\bf \:  =  \: - \: \dfrac{1}{12}$

Hence,

$\red{\rm :\longmapsto\:\boxed{ \rm \:\displaystyle\lim_{x \to 0} \frac{ \sqrt{cosx} - \sqrt[3]{cosx} }{{sin}^{2}x } = \: - \: \frac{1}{12} }}$

Additional Information :-

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

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

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

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

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