Question:
If the function f(x) satisfies \(\lim_{x\rightarrow 1}\) \(\frac{f(x)-2}{x^2-1}\) =\(\pi\), evaluate \(\lim_{x\rightarrow 1}\) f(x).
If the function f(x) satisfies \(\lim_{x\rightarrow 1}\) \(\frac{f(x)-2}{x^2-1}\) =\(\pi\), evaluate \(\lim_{x\rightarrow 1}\) f(x).
Updated On: Dec 14, 2025
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Solution and Explanation
\(\lim_{x\rightarrow 1}\) \(\frac{f(x)-2}{x^2-1}\) =\(\pi\)
\(\Rightarrow\)\(\frac{\lim_{x\rightarrow 1} (f(x)-2)}{\lim_{x\rightarrow 1}(x^2-1)}\) =\(\pi\)
\(\Rightarrow\)\(\lim_{x\rightarrow 1}\) (f(x)-2)=\(\pi\) \(\lim_{x\rightarrow 1}\) (x2-1)
\(\Rightarrow\)\(\lim_{x\rightarrow 1}\) (f(x)-2)=\(\pi\)(12-1)
\(\Rightarrow\)\(\lim_{x\rightarrow 1}\) (f(x)-2)=0
\(\Rightarrow\)\(\lim_{x\rightarrow 1}\) f(x)-\(\lim_{x\rightarrow 1}\)2=0
\(\Rightarrow\)\(\lim_{x\rightarrow 1}\) f(x)- 2=0
∴\(\lim_{x\rightarrow 1}\) f(x) = 2
\(\Rightarrow\)\(\frac{\lim_{x\rightarrow 1} (f(x)-2)}{\lim_{x\rightarrow 1}(x^2-1)}\) =\(\pi\)
\(\Rightarrow\)\(\lim_{x\rightarrow 1}\) (f(x)-2)=\(\pi\) \(\lim_{x\rightarrow 1}\) (x2-1)
\(\Rightarrow\)\(\lim_{x\rightarrow 1}\) (f(x)-2)=\(\pi\)(12-1)
\(\Rightarrow\)\(\lim_{x\rightarrow 1}\) (f(x)-2)=0
\(\Rightarrow\)\(\lim_{x\rightarrow 1}\) f(x)-\(\lim_{x\rightarrow 1}\)2=0
\(\Rightarrow\)\(\lim_{x\rightarrow 1}\) f(x)- 2=0
∴\(\lim_{x\rightarrow 1}\) f(x) = 2
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