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Question
\[f\left( x \right) = \frac{a x^2 + b}{x^2 + 1}, \lim_{x \to 0} f\left( x \right) = 1\] and \[\lim_{x \to \infty} f\left( x \right) = 1,\]then prove that f(−2) = f(2) = 1
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Solution
\[f\left( x \right) = \frac{a x^2 + b}{x^2 + 1}\]
\[ \lim_{x \to 0} f\left( x \right) = 1\]
\[ \Rightarrow \lim_{x \to 0} \left[ \frac{a x^2 + b}{x^2 + 1} \right] = 1\]
\[ \Rightarrow \frac{a \times 0 + b}{a + 1} = 1\]
\[ \Rightarrow b = 1\]
\[\text{ Also }, \lim_{x \to \infty} f\left( x \right) = 1\]
\[ \lim_{x \to \infty} \left[ \frac{a x^2 + b}{x^2 + 1} \right] = 1\]Dividing the numerator and the denominator by x2:
\[\Rightarrow \lim_{x \to \infty} \left[ \frac{a + \frac{b}{x^2}}{1 + \frac{1}{x^2}} \right] = 1\]
\[As x \to \infty , \frac{1}{x}, \frac{1}{x^2} \to 0\]
\[ \Rightarrow \frac{a + 0}{1 + 0} = 1\]
\[ \Rightarrow a = 1\]
\[ \therefore a = 1, b = 1\]
\[ \Rightarrow f\left( x \right) = \frac{x^2 + 1}{x^2 + 1} = 1\]
\[f\left( - 2 \right) = 1 \left[ \text{ Since f }\left( x \right) \text{ is a constant function, its value does not depend on x } . \right]\]
\[f\left( 2 \right) = 1\]
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