Question

${lim}_{x\to \pi /4}\frac{\sqrt{2}\cos x-1}{\cot x-1}$ is equal to

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{2}$
• C. $\frac{1}{2\sqrt{2}}$
• D. $1$

Answer 1

Answer: (B)

$\frac{1}{2}$

let $\underset{x\to \frac{x}{4}}{lim}\frac{\sqrt{2}cosx-1}{cotx-1}=p$

Applying L - hospital rule, since it is of the from $\frac{0}{0}$

$P=\underset{x\to \frac{x}{4}}{lim}\frac{+\sqrt{2}sinx}{+cose{c}^{2}x}$

$P=\frac{-\sqrt{2}\times \frac{1}{\sqrt{2}}}{-(\sqrt{2}{)}^2}\Rightarrow \frac{-1}{-2}=\frac{1}{2}$

$\therefore$ limit is $\frac{1}{2}.$ so option (b) is correct.