Question

Solve the given limit without using L'Hopital rule.

$\rm\lim\limits_{x\to0}\dfrac{sin^2(\pi\cos^4x)}{x^4}$


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Answer 1

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

Given expression is

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

If we substitute directly x = 0, we get

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

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

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

which is indeterminant form.

Again, Consider

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

can be rewritten as

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

We know,

$\boxed{ \tt{ \: {sin}^{2}x + {cos}^{2}x = 1 \: }}$

So, we get

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

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

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

We know

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: sin(\pi + x) = - sinx \: }}}$

So, we get

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

$\boxed{ \tt{ \: \sf \: as \: x \: \to \: 0 \: so, \: {sin}^{4}x - {2sin}^{2}x \: \to \: 0}}$

Thus,

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

We know

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

So, using this, we get

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

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

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

We know,

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

So, using this, we get

$\rm \: = \: {\pi}^{2} \times 1 \times {(0 - 2)}^{2}$

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

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

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

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

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

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

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

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

Answer 2

Answer:

4π²

Step-by-step explanation:

Given :

$\mathrm{\lim \limits_{x \rightarrow 0} \dfrac{sin^2(\pi cos^4x)}{x^4}}$

To find :

Evaluate the given limit

Solution :

First let us substitute 0 and check the result

$\longmapsto \mathrm{\lim \limits_{x \rightarrow 0} \dfrac{sin^2(\pi cos^4x)}{x^4}}$

$\implies \mathrm{\dfrac{sin^2(\pi cos^4(0))}{0^4}}$

$\implies \mathrm{\dfrac{sin^2(\pi)}{0}}$

$\implies \mathrm{\dfrac{0}{0}}$

Hence, it is in the 0/0 form which is indeterminate form

First let us find what is cos⁴x

$\implies \mathrm{cos^4x}$

We know that,

$\boxed{\mathrm{sin^2x + cos^2x = 1}}$

$\implies \mathrm{(cos^2x)^2}$

$\implies \mathrm{(1 - sin^2x)^2}$

$\implies \mathrm{(1 - 2sin^2x + sin^4x)}$

So, substituting in place of cos⁴x in the given equation,

$\implies \mathrm{\lim \limits_{x \rightarrow 0} \dfrac{sin^2(\pi \mathrm{(1 - 2sin^2x + sin^4x)})}{x^4}}$

$\implies \mathrm{\lim \limits_{x \rightarrow 0} \dfrac{sin^2(\pi + \pi \mathrm{(sin^4x - 2sin^2x)})}{x^4}}$

We know that,

$\boxed{\mathrm{sin(\pi + x) = -sinx }}$

$\implies \mathrm{\lim \limits_{x \rightarrow 0} \dfrac{\big(-sin(\pi \mathrm{(sin^4x - 2sin^2x)})\big)^2}{x^4}}$

$\implies \mathrm{\lim \limits_{x \rightarrow 0} \dfrac{sin^2(\pi \mathrm{(sin^4x - 2sin^2x)})}{x^4}}$

Now, multiplying and dividing with $\mathrm{\pi^2(sin^4x - 2sin^2x)^2}$ as there is numerator with power 2, we get,

$\implies \mathrm{\lim \limits_{x \rightarrow 0} \dfrac{sin^2(\pi \mathrm{(sin^4x - 2sin^2x)})}{\mathrm{\pi^2(sin^4x - 2sin^2x)^2}} \times \dfrac{\mathrm{\pi^2(sin^4x - 2sin^2x)^2}}{x^4}}$

We know that,

$\boxed{\mathrm{\lim \limits _{x \rightarrow 0} \dfrac{sinx}{x} = 1}}$

$\boxed{\mathrm{(a -b)^2 = a^2 - 2ab + b^2 }}$

So, applying this in above equation,

$\implies \mathrm{\lim \limits_{x \rightarrow 0} \dfrac{\mathrm{\pi^2(sin^8x - 4sin^6x + 4sin^4x)}}{x^4}}$

Taking sin⁴x common,

$\implies \mathrm{\lim \limits_{x \rightarrow 0} \dfrac{\pi^2\mathrm{sin^4x(sin^4x - 4sin^2x + 4)}}{x^4}}$

$\implies \mathrm{\lim \limits_{x \rightarrow 0} \pi^2 \times \dfrac{\mathrm{sin^4x}}{x^4} \times (sin^4x - 4sin^2x + 4)}$

$\implies \mathrm{ \pi ^2 \times 1 \times (sin0 - 4sin0 + 4)}$

We know that,

$\boxed{\mathrm {sin 0 = 0}}$

$\implies \mathrm{ \pi^2 \times (4)}$

$\implies \mathrm{ 4\pi^2}$

$\therefore \underline{\boxed{\mathbf{\lim \limits_{x \rightarrow 0} \dfrac{sin^2(\pi cos^4x)}{x^4} = 4\pi^2}}}$

Hope it helps!!