Question

$\underset{x\to \pi /4}{lim}\left(\frac{1-tanx}{1-\sqrt{2}sinx}\right)$ is equal to

• A. 0
• B. 1
• C. -2
• D. 2

Answer 1

The correct option is D 2

$\underset{x\to \pi /4}{lim}\left(\frac{1-tanx}{1-\sqrt{2}sinx}\right)$

$=\underset{x\to \pi /4}{lim}\left(\frac{1-tanx}{1-\sqrt{2}sinx}\right).\frac{1+\sqrt{2}sinx}{1+\sqrt{2}sinx}$

$=\underset{x\to \pi /4}{lim}\left(\frac{1-tanx}{1-2si{n}^{2}x}\right).\frac{1+\sqrt{2}sinx}{1}$

$=\underset{x\to \pi /4}{lim}\left(\frac{1-\sin x/\cos x}{{\cos }^{2}x-si{n}^{2}x}\right).\frac{1+\sqrt{2}sinx}{1}$

$=\underset{x\to \pi /4}{lim}\left(\frac{1}{\cos x(\cos x+sinx)}\right).\frac{1+\sqrt{2}sinx}{1}=2$