Question
Evaluate the following limits:
$\lim _{x \rightarrow 0}\left[\frac{\sec x^2-1}{x^4}\right]$

Answer

$\lim _{x \rightarrow 0} \frac{\sec x^2-1}{x^4}$
Put $x^2=y$
As $x \rightarrow 0, x^2 \rightarrow 0$
$\therefore \quad y \rightarrow 0$
$\therefore \quad$ Required limit
$ \quad=\lim _{y \rightarrow 0} \frac{\sec y-1}{y^2}$
$\quad=\lim _{y \rightarrow 0} \frac{\sec y-1}{y^2} \times \frac{\sec y+1}{\sec y+1}$
$\quad=\lim _{y \rightarrow 0} \frac{\sec ^2 y-1}{y^2(\sec y+1)}$
$=\lim _{y \rightarrow 0} \frac{\tan ^2 y}{y^2(\sec y+1)}$
$=\lim _{y \rightarrow 0} \frac{\tan ^2 y}{y^2} \times \frac{1}{\sec y+1}$
$=\lim _{y \rightarrow 0}\left(\frac{\tan ^2 y}{y^2}\right) \times \lim _{y \rightarrow 0} \frac{1}{\sec y+1}$
$=(1)^2 \times \frac{1}{1+1}$
$=\frac{1}{2} $

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