Given: \(\cfrac{a\text x^2+b}{\text x^2+1},\) \(\lim\limits_{\text x \to 0} \) f(x) = 1, and \(\lim\limits_{\text x \to \infty} \) f(x) = 1
To Prove: f(-2) = f(2) = 1
Proof: we have, f(x) = \(\cfrac{a\text x^2+b}{\text x^2+1}\)
And, \(\lim\limits_{\text x \to 0} \) f(x) = 1
b = 1
Thus, f(x) = \(\cfrac{a\text x^2+b}{\text x^2+1}\)
On substituting the value of a and b we get,
So, f(x) = 1
Then, f(-2) = 1
Also, f(2) = 1
Hence, f(2) = f(-2) = 1