Evaluate the given limit $$\lim\limits_{x \to \dfrac{\pi}{2}}\dfrac{tan2x}{x-\dfrac{\pi}{2}}$$

Asked by Abhisek | 1 year ago |  69

##### Solution :-

$$\lim\limits_{x \to \dfrac{\pi}{2}}\dfrac{tan2x}{x-\dfrac{\pi}{2}}$$

At $$x=\dfrac{\pi}{2}$$ the value of the given function takes the form $$\dfrac{0}{0}$$

Now, put So that $$x-\dfrac{\pi}{2}=y$$ so that $$x\rightarrow \dfrac{\pi}{2},y\rightarrow 0$$

$$\lim\limits_{y \to 0}\dfrac{(y+\dfrac{\pi}{2})}{y}$$

$$\lim\limits_{y \to 0}\dfrac{tan2y}{y}$$

$$\lim\limits_{y \to 0}\dfrac{sin2y}{cos2y}$$

$$( \lim\limits_{y \to 0}\dfrac{sin2y}{2y})\times ( \lim\limits_{y \to 0}\times \dfrac{2}{cos2y})$$

$$1\times \dfrac{2}{cos0}$$

$$1\times \dfrac{2}{1}$$

= 2

Answered by Pragya Singh | 1 year ago

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