12 December 2021

Statement

Find the value of the following limit.

\(\lim\limits_{x \to \infty} \frac{\sqrt{1+4x^{2}}}{4 + x}\)

Solution

\(\lim\limits_{x \to \infty} \frac{\sqrt{1+4x^{2}}}{4 + x}\)

Since we want to find the value of the function when \(x \to +\infty\) it won't be a problem if we remove the \(x^2\) from the square root. Note that if we were asked to find the value for some negative \(x\) this operation wouldn't be valid since \(\sqrt{x^2} \neq x\) for \(x < 0\)

\(= \lim\limits_{x \to \infty} \frac{\sqrt{x^2(4 + \frac{1}{x^2})}}{4 + x} \)

\(= \lim\limits_{x \to \infty} \frac{x\sqrt{4 + \frac{1}{x^2}}}{x(1 + \frac{4}{x})} \)

\(= \lim\limits_{x \to \infty} \frac{\sqrt{4 + \frac{1}{x^2}}}{1 + \frac{4}{x}} \)

Since the denominator \(1 + \frac{4}{x} \neq 0\) when \(x \to \infty\) we can apply the following property.

\(= \frac{\lim\limits_{x \to \infty} \sqrt{4 + \frac{1}{x^2}}}{\lim\limits_{x \to \infty} 1 + \frac{4}{x}} \)

\(= \frac{\sqrt{4}}{1} \)

\(= 2 \)