Question

If

$\lim \: x \to \: 0 \: \frac{sin(\pi \: {cos}^{2}x) }{ k{x}^{2} } = 1$

find the value of k​

Answer 1

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

Given that,

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

If we substitute directly x = 0, we get

$\rm \: \frac{sin(\pi \: {cos}^{2} 0)}{ {k \: 0}^{2} } = 1$

$\rm \: \frac{sin\pi }{0} = 1$

$\rm \: \frac{0 }{0} = 1$

which is indeterminant form.

So, Consider again

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

can be rewritten as

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

$\rm \: \displaystyle\lim_{x \to 0}\rm \frac{sin[\pi - \pi{sin}^{2} x]}{ {k \: x}^{2} } = 1$

We know,

$\boxed{\sf{  \:sin(\pi - x) \: = \: sinx \: }}$

So, using this result, we get

$\rm \: \displaystyle\lim_{x \to 0}\rm \frac{sin[\pi{sin}^{2} x]}{ {k \: x}^{2} } = 1$

$\rm \: \displaystyle\lim_{x \to 0}\rm \frac{sin[\pi{sin}^{2} x]}{\pi \: {sin}^{2}x } \times \frac{\pi \: {sin}^{2} x}{k \: {x}^{2} } = 1$

We know,

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

So, using this result, we get

$\rm \: {1}^{2} \times \pi \times \dfrac{ {1}^{2} }{k} = 1$

$\rm \: \dfrac{\pi}{k} = 1$

$\rm\implies \:k \: = \: \pi \:$

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Additional Information :-

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

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

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

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

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