Question

${lim}_{x\to \frac{\pi }{4}}\frac{\sqrt{2}cosx-1}{cotx-1}$ is equal to

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

Answer 1

The correct option is B

$\frac{1}{2}$

${lim}_{x\to \frac{\pi }{4}}\frac{\sqrt{2}cosx-1}{cotx-1}$
Rationalising the numerator, we get:
$={lim}_{x\to \frac{\pi }{4}}\left(\frac{\sqrt{2}cosx-1}{cotx-1}\right)\times \left(\frac{\sqrt{2}cosx+1}{\sqrt{2}cosx+1}\right)$
$={lim}_{x\to \frac{\pi }{4}}\frac{(2co{s}^{2}x-1)}{(cosx-sinx)}\times \frac{sinx}{(\sqrt{2}cosx+1)}$
${lim}_{x\to \frac{\pi }{4}}\frac{(co{s}^{2}x-si{n}^{2}x)}{(cosx-sinx)}\times \frac{sinx}{(\sqrt{2}cosx+1)}$
$={lim}_{x\to \frac{\pi }{4}}\frac{(cosx+sinx)sinx}{\left(\sqrt{2cosx+1}\right)}$
$=\frac{(cos\frac{\pi }{4}+sin\frac{\pi }{4})sin\frac{\pi }{4}}{(\sqrt{2}cos\frac{\pi }{4}+1)}$
$={lim}_{x\to \frac{\pi }{4}}=\frac{\left(\frac{1}{\sqrt{2}+\frac{1}{\sqrt{2}}}\right)\left(\frac{1}{\sqrt{2}}\right)}{\sqrt{2}.\frac{1}{2}+1}$
$=\frac{(\frac{2}{\sqrt{2}}\times \frac{1}{\sqrt{2}})}{2}$
$=\frac{1}{2}$