Question

Find the value of $\underset{x\to \frac{\pi }{4}}{lim}\frac{cosx-sinx}{cos2x}$

• A.
• B.
• C.
• D. limit does not exist

Answer 1

The correct option is B

Lets substitute $x=\frac{\pi }{4}$ to see if we can solve it by direct substitution. We get $\frac{\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}}{0}=\frac{0}{0}$. We get the

indeterminate form again. If there is a common factor, we can cancel that.
For that we can write

$cos2xasco{s}^{2}x-si{n}^{2}x$. We do this because there is $cosx-sinx$ in the numerator.

$\Rightarrow \underset{x\to \frac{\pi }{4}}{lim}\frac{cosx-sinx}{cos2x}={lim}_{x\to \frac{\pi }{4}}\frac{cosx-sinx}{co{s}^{2}x-si{n}^{2}x}$

$=\underset{x\to \frac{\pi }{4}}{lim}\frac{cosx-sinx}{(cosx+sinx)(cosx-sinx)}$

Since $x\ne \frac{\pi }{4}$
$cosx-sinx\ne 0orcosx\ne sinx$ we can write above expression as -
$=\underset{x\to \frac{\pi }{4}}{lim}\frac{1}{cosx+sinx}$

$=\frac{1}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}=\frac{1}{\sqrt{2}}$