GATE (TF) Textile 2020 Question Paper Solution | GATE/2020/TF/26

Let . The value of L is $L= \lim_{x\rightarrow \frac{\pi}{2}} (sinx)^{tanx}$ . The value of $L$ is ________.

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$L= \lim_{x\rightarrow \frac{\pi}{2}} (sinx)^{tanx}$

If $L= \lim (f(x))^{g(x)}$

Then, $L= e^\lim (f(x))^{g(x)}$

$L= e^\lim (f(x)-1)^{g(x)}$

$L= e^\lim_{x\rightarrow \frac{\pi}{2}} (sinx-1)^{tanx}$

$L= e^\lim_{x\rightarrow \frac{\pi}{2}} (sinx-1)(tanx)$

This is the case of $\frac{\infinity}{\infinity}$

Multiply and devide this by (sinx+1)

$L= e^\lim_{x\rightarrow \frac{\pi}{2}} \frac{sinx^2-1}{sinx+1} \times tanx$

$L= e^\lim_{x\rightarrow \frac{\pi}{2}} \frac{cos^2 x}{sinx+1} \times tanx$

$L= e^\lim_{x\rightarrow \frac{\pi}{2}} \frac{cos^2 x}{sinx+1} \times \frac{sinx}{cosx}$

$L= e^\lim_{x\rightarrow \frac{\pi}{2}} \frac{cos x \times sin x}{sinx+1}$

Multiply and devide by 2

$L= e^\lim_{x\rightarrow \frac{\pi}{2}} \frac{2cos x \times sin x}{2(sinx+1)}$

$L= e^\lim_{x\rightarrow \frac{\pi}{2}} \frac{sin2x}{2(sinx+1)}$

By taking limit x- $\frac{\pi}{2}$

$L=e^\lim_{x\rightarrow \frac{\pi}{2}} \frac{sin \pi}{2(sin\frac{\pi}{2}+1)}$

$L=e^0$

L=1 (Ans)