Question

$\underset{x\to \frac{\pi }{4}}{lim}\frac{1-\tan x}{1-\sqrt{2}\sin x}$.

Answer 1

Given limit is $\underset{x\to \frac{\pi }{4}}{lim}\frac{1-\tan x}{1-\sqrt{2}\sin x}$

We can clearly observe that the given limit is in $\frac{0}{0}$ form,

So, we can apply L-hospitals rule

$\Rightarrow \underset{x\to \frac{\pi }{4}}{lim}\frac{1-\tan x}{1-\sqrt{2}\sin x}$

$=\underset{x\to \frac{\pi }{4}}{lim}\frac{1-\tan x}{1-\sqrt{2}\sin x}$

$=\underset{x\to \frac{\pi }{4}}{lim}\frac{{\sec }^{2}x}{\sqrt{2}\cos x}$

$=\frac{(\sqrt{2}{)}^2}{\sqrt{2}.\frac{1}{\sqrt{2}}}=2$