Question

$\large \displaystyle \tt \color{green}{\lim_{x \to0}} \bigg[ \frac{1}{ {x}^{2} } - \frac{1}{(x \cos( \alpha x))^{2} } = - 9\bigg]$


Find the value of alpha​​

Answer 1

Answer:

Step-by-step explanation:

$\large \displaystyle \tt \lim_{x \to 0}} \bigg[ \frac{1}{ {x}^{2} } - \frac{1}{(xcos(\alpha x))^2} = - 9\bigg]$

$or, \large \displaystyle \tt \lim_{x \to 0}} \bigg[ \frac{1}{ {x}^{2} } - \frac{1}{x^2cos^{2}(\alpha x)} = - 9\bigg]\\or, \large \displaystyle \tt \lim_{x \to 0}} \frac{1}{x^2} \bigg[1- \frac{1}{cos^{2}(\alpha x)} = - 9\bigg]\\or, \large \displaystyle \tt \lim_{x \to 0}} \frac{1}{x^2} \bigg[\frac{cos^{2}{(\alpha x)}-1}{cos^{2}(\alpha x)} = - 9\bigg]\\or, \large \displaystyle \tt \lim_{x \to 0}} \frac{1}{x^2} \bigg[\frac{1-cos^{2}{(\alpha x)}}{cos^{2}(\alpha x)} = 9\bigg]$

$or, \large \displaystyle \tt \lim_{x \to 0}} \frac{1}{x^2} \bigg[\frac{sin^{2}{(\alpha x)}}{cos^{2}(\alpha x)} = 9\bigg]\\or, \large \displaystyle \tt \lim_{x \to 0}} \bigg[\frac{sin^{2}{(\alpha x)}}{(\alpha x)^2}\cdot \frac{\alpha^2}{cos^{2}{\alpha x}} = 9\bigg]\\or, \bigg(\large \displaystyle \tt \lim_{x \to 0 } \frac{sin(\alpha x)}{\alpha x} \bigg)^2\cdot \frac{\alpha^2}{\bigg(\large \displaystyle \tt \lim_{x \to 0 } (cos \alpha x) \bigg)^2} = 9\\ or, (1)^2 . \frac{\alpha^2}{(1)^2} = 9$$or, \alpha^2 = 9~~ (\because \lim_{x \to 0} \frac{\sin{x}}{x} = 1\text{ and } \lim_{x \to 0} \cos{x} = 1~)\\or, \alpha = \pm3$

Answer 2

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

Given expression is

$\rm :\longmapsto\:\displaystyle \tt \color{green}{\lim_{x \to0}} \bigg[ \frac{1}{ {x}^{2} } - \frac{1}{(x \cos( \alpha x))^{2} }\bigg] = - 9$

can be rewritten as

$\rm :\longmapsto\:\displaystyle \rm {\lim_{x \to0}} \bigg[ \frac{1}{ {x}^{2} } - \frac{1}{ {x}^{2} \cos^{2} \alpha x }\bigg] = - 9$

$\rm :\longmapsto\:\displaystyle \rm {\lim_{x \to0}} \bigg[ \frac{ {cos}^{2} \alpha x - 1}{ {x}^{2} \cos^{2} \alpha x }\bigg] = - 9$

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{1}{ {cos}^{2} \alpha x} \times \displaystyle\lim_{x \to 0}\rm \frac{ - (1 - {cos}^{2} \alpha x)}{ {x}^{2} } = - 9$

We know,

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

So, using this identity, we get

$\rm :\longmapsto\: \dfrac{1}{ {cos}^{2}0} \times \displaystyle\lim_{x \to 0}\rm \frac{{sin}^{2}\alpha x}{ {x}^{2} } = 9$

$\rm :\longmapsto\: 1\times \displaystyle\lim_{x \to 0}\rm \frac{{sin}^{2}\alpha x}{ { \alpha }^{2} {x}^{2} } \times { \alpha }^{2} = 9$

We know,

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

So, using this identity, we get

$\rm :\longmapsto\: {1}^{2} \times { \alpha }^{2} = 9$

$\rm :\longmapsto\:{\alpha }^{2} = 9$

$\rm :\longmapsto\:{\alpha }^{2} = {3}^{2}$

$\bf\implies \: \alpha \: = \: \pm \: 3$

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ADDITIONAL INFORMATION

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

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

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

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

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