Question

Evaluate the given limits :-
$\rm\lim\limits_{x\to0}\dfrac{sin^2(\pi\cos^4x)}{x^4}$
​

Answer 1

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

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\sf \frac{ {sin}^{2}(\pi \: {cos}^{4}x)}{ {x}^{4} }$

If we substitute directly x = 0, we get

$\rm \: = \: \sf \dfrac{ {sin}^{2}(\pi \: {cos}^{4}0)}{ {0}^{4} }$

$\rm \: = \:\sf \dfrac{ {sin}^{2}(\pi \: \times 1)}{ 0 }$

$\rm \: = \:\sf \dfrac{ {sin}^{2}\pi }{ 0 }$

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

which is indeterminant form.

So, Consider Again

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\sf \frac{ {sin}^{2}(\pi \: {cos}^{4}x)}{ {x}^{4} }$

Using L - Hospital Rule, we have

$\rm \: = \: \displaystyle\lim_{x \to 0}\sf \frac{ \dfrac{d}{dx}{sin}^{2}(\pi \: {cos}^{4}x)}{ \dfrac{d}{dx}{x}^{4} }$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: }}$

and

$\boxed{ \tt{ \: \dfrac{d}{dx}sinx = cosx \: }}$

So, using this, we get

$\rm \: = \: \displaystyle\lim_{x \to 0}\sf \dfrac{2sin(\pi {cos}^{4}x)\dfrac{d}{dx}sin(\pi \: {cos}^{4}x) }{4 {x}^{3} }$

$\rm \: = \: \displaystyle\lim_{x \to 0}\sf \dfrac{2sin(\pi {cos}^{4}x)cos(\pi {cos}^{4}x) \dfrac{d}{dx}(\pi \: {cos}^{4}x) }{4 {x}^{3} }$

$\rm \: = \: \displaystyle\lim_{x \to 0}\sf \dfrac{2sin(\pi {cos}^{4}x)cos(\pi {cos}^{4}x) (4\pi{cos}^{3}x) ( - sinx)}{4 {x}^{3} }$

$\rm \: = \: \displaystyle\lim_{x \to 0}\sf \dfrac{2sin(\pi {cos}^{4}x)cos(\pi {cos}^{4}x) (\pi{cos}^{3}x) ( - sinx)}{{x}^{3} }$

$\rm \: = - \: 2\pi \: cos\pi \: \displaystyle\lim_{x \to 0}\sf \frac{sin(\pi {cos}^{4} x)}{ {x}^{2} } \times \displaystyle\lim_{x \to 0}\sf \frac{sinx}{x}$

We know,

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

So, using this, we get

$\rm \: = - \: 2\pi \: ( - 1) \: \displaystyle\lim_{x \to 0}\sf \frac{sin(\pi {cos}^{4} x)}{ {x}^{2} } \times 1$

$\rm \: = \: 2\pi \: \displaystyle\lim_{x \to 0}\sf \frac{sin(\pi {cos}^{4} x)}{ {x}^{2} }$

Using L - Hospital Rule, we get

$\rm \: = \: 2\pi \: \displaystyle\lim_{x \to 0}\sf \frac{\dfrac{d}{dx}sin(\pi {cos}^{4} x)}{ \dfrac{d}{dx}{x}^{2} }$

$\rm \: = \: 2\pi \: \displaystyle\lim_{x \to 0}\sf \frac{cos(\pi {cos}^{4}x) \dfrac{d}{dx}\pi {cos}^{4} x}{2x }$

$\rm \: = \: \pi \: \displaystyle\lim_{x \to 0}\sf \frac{cos(\pi {cos}^{4}x) \pi (4{cos}^{3} x)( - sinx)}{x }$

$\rm \: = \: - \: 4 {\pi}^{2} cos(\pi) \: \displaystyle\lim_{x \to 0}\sf \frac{sinx}{x }$

$\rm \: = \: - \: 4 {\pi}^{2} \times ( - 1) \: \times 1$

$\rm \: = \: \: 4 {\pi}^{2}$

Hence,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 0}\sf \frac{ {sin}^{2}(\pi \: {cos}^{4}x)}{ {x}^{4} } = {4\pi}^{2} \: }}}$

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Additional Information :-

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

Answer 2

Given to evaluate the limit,

$\small\displaystyle\longrightarrow L=\lim_{x\to0}\dfrac{\sin^2(\pi\cos^4x)}{x^4}$

On putting x=0 directly then we get indeterminate form.

We know that,

• $\small\displaystyle\sin(\pi-x)=\sin x$

Thus we have,

• $\small\displaystyle\sin(\pi-\pi\cos^4x)=\sin(\pi\cos^4x)$

Then,

$\small\displaystyle\longrightarrow L=\lim_{x\to0}\dfrac{\sin^2(\pi-\pi\cos^4x)}{x^4}$

Dividing numerator and denominator by $\small(\pi-\pi\cos^4x)^2,$

$\small\text{ \displaystyle\longrightarrow L=\lim_{x\to0}\dfrac{\left(\dfrac{\sin^2(\pi-\pi\cos^4x)}{(\pi-\pi\cos^4x)^2}\right)}{\left(\dfrac{x^4}{(\pi-\pi\cos^4x)^2}\right)} }$

Since $\small\text{ \dfrac{\left(\dfrac{a}{b}\right)}{\left(\dfrac{c}{d}\right)}=\dfrac{a}{b}\cdot\dfrac{d}{c}, }$

$\small\displaystyle\longrightarrow L=\lim_{x\to0}\dfrac{\sin^2(\pi-\pi\cos^4x)}{(\pi-\pi\cos^4x)^2}\cdot\dfrac{(\pi-\pi\cos^4x)^2}{x^4}$

$\small\displaystyle\longrightarrow L=\lim_{x\to0}\left(\dfrac{\sin(\pi-\pi\cos^4x)}{\pi-\pi\cos^4x}\right)^2\cdot\lim_{x\to0}\dfrac{(\pi-\pi\cos^4x)^2}{x^4}$

$\small\displaystyle\longrightarrow L=\left(\lim_{x\to0}\dfrac{\sin(\pi-\pi\cos^4x)}{\pi-\pi\cos^4x}\right)^2\lim_{x\to0}\dfrac{(\pi-\pi\cos^4x)^2}{x^4}$

As $\small\pi-\pi\cos^4x\to0$ when $\smallx\to0,$

$\small\displaystyle\longrightarrow L=\left(\lim_{\pi-\pi\cos^4x\to0}\dfrac{\sin(\pi-\pi\cos^4x)}{\pi-\pi\cos^4x}\right)^2\lim_{x\to0}\dfrac{(\pi-\pi\cos^4x)^2}{x^4}$

Taking $\smallt=\pi-\pi\cos^4x,$

$\small\displaystyle\longrightarrow L=\left(\lim_{t\to0}\dfrac{\sin t}{t}\right)^2\lim_{x\to0}\dfrac{(\pi-\pi\cos^4x)^2}{x^4}$

We know that,

• $\small\displaystyle\lim_{t\to0}\dfrac{\sin t}{t}=1$

Then,

$\small\displaystyle\longrightarrow L=1^2\cdot\lim_{x\to0}\dfrac{(\pi-\pi\cos^4x)^2}{x^4}$

$\small\displaystyle\longrightarrow L=\lim_{x\to0}\dfrac{\pi^2(1-\cos^4x)^2}{x^4}$

$\small\displaystyle\longrightarrow L=\pi^2\lim_{x\to0}\left(\dfrac{1-\cos^4x}{x^2}\right)^2$

$\small\displaystyle\longrightarrow L=\pi^2\left(\lim_{x\to0}\dfrac{1-\cos^4x}{x^2}\right)^2$

$\small\displaystyle\longrightarrow L=\pi^2\left(\lim_{x\to0}\dfrac{(1-\cos^2x)(1+\cos^2x)}{x^2}\right)^2$

$\small\displaystyle\longrightarrow L=\pi^2\left(\lim_{x\to0}\dfrac{\sin^2x(1+\cos^2x)}{x^2}\right)^2$

$\small\displaystyle\longrightarrow L=\pi^2\left(\lim_{x\to0}\dfrac{\sin^2x}{x^2}\cdot\lim_{x\to0}(1+\cos^2x)\right)^2$

$\small\displaystyle\longrightarrow L=\pi^2\left(\left(\lim_{x\to0}\dfrac{\sin x}{x}\right)^2\cdot\lim_{x\to0}(1+\cos^2x)\right)^2$

$\small\displaystyle\longrightarrow L=\pi^2\left(1^2(1+\cos^20)\right)^2$

$\small\displaystyle\longrightarrow L=\pi^2\left(1^2(1+1^2)\right)^2$

$\small\text{ \displaystyle\longrightarrow\underline{\underline{L=4\pi^2}} }$