Question

If $f\left(x\right)=\frac{\sqrt{2}\cos x-1}{\cot x-1},x\ne \frac{\pi }{4}$, find the value of $f\left(\frac{\pi }{4}\right)$ so that function f becomes continuous at $x=\frac{\pi }{4}$.

Answer 1

$\begin{array}{c}f\left(x\right)=\frac{\sqrt{2}\cos x-1}{\cot +x-1}\\ f\left(\frac{\pi }{4}\right)=2\\ f\left(x\right)=\frac{\sqrt{2}\cos x-1}{\frac{\cos x}{\sin x}-1}\\ \frac{\left(\sqrt{2}\cos z-1\right)\sin x}{\cos x-\sin x}\\ u\sin gLhospitalrule\\ =\frac{\sqrt{2}\sin x-0}{-\cos ecx}\\ =\frac{\sqrt{2}\sin x}{-\cos e{c}^{2}x}\\ f\left(\frac{\pi }{4}\right)=\frac{\sqrt{2}\sin \left(\frac{\pi }{4}\right)}{\cos e{c}^2\left(\frac{\pi }{4}\right)}\\ =\frac{\sqrt{2}\times \frac{1}{\sqrt{2}}}{{\left(\sqrt{2}\right)}^2}\\ \frac{1}{2}\end{array}$