Evaluate the following limit : f(x) = (ax^2 + b)/(x^2 + 1), lim(x→0)f(x) = 1 and lim(x→∞)f(x) = 1, then prove that f(-2) = f(2) = 1.

Question

Evaluate the following limit : f(x) = \(\cfrac{a\text x^2+b}{\text x^2+1},\)\(\lim\limits_{\text x \to 0} \) f(x) = 1, and \(\lim\limits_{\text x \to \infty} \) f(x) = 1,

then prove that f(-2) = f(2) = 1.

f(x) = (ax² + b)/(x² + 1), lim(x→0)f(x) = 1 and lim(x→∞)f(x) = 1,

Answer 1

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