Question

# If the function$$f(x)$$ satisfies $$\displaystyle \lim_{x\rightarrow 1}\frac{f\left ( x \right )-2}{x^{2}-1}=\pi$$ evaluate $$\displaystyle \lim_{x\rightarrow 1}f\left ( x \right )$$

Solution

## $$\displaystyle \lim_{x\rightarrow 1}\frac{f\left ( x \right )-2}{x^{2}-1}=\pi$$$$\displaystyle \Rightarrow \frac{\lim_{x\rightarrow 1}\left ( f\left ( x \right )-2 \right )}{\lim_{x\rightarrow 1}\left ( x^{2} -1\right )}=\pi$$$$\displaystyle \Rightarrow \lim_{x\rightarrow 1}\left ( f\left ( x \right )-2 \right )=\pi \lim_{x\rightarrow 1}\left ( x^{2}-1 \right )$$$$\displaystyle \Rightarrow \lim_{x\rightarrow 1}\left ( f\left ( x \right )-2 \right )=\pi \left ( 1^{2}-1 \right )$$$$\displaystyle \Rightarrow \lim_{x\rightarrow 1}\left ( f\left ( x \right )-2 \right )=0$$$$\displaystyle \Rightarrow \lim_{x\rightarrow 1}f\left ( x \right )-\lim_{x\rightarrow 1}2=0$$$$\displaystyle \Rightarrow \lim_{x\rightarrow 1}f\left ( x \right )-2=0$$$$\displaystyle \therefore \lim_{x\rightarrow 1}f\left ( x \right )=2$$MathematicsNCERTStandard XI

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