Question

My teachers key says that it us -2. I was never taught how to evaluate limits that go to infinity or negative infinity so I need to be taught how to solve them.

Answer 1

SOLUTION

The given limit is:

$\lim_{x\to-\infty}\frac{\sqrt{4x^2+5}}{x-3}$

Divide by the numerator and denominator by x

$\operatorname{\lim}_{x\to-\infty}\frac{\frac{\sqrt{4x^2+5}}{x}}{1-\frac{3}{x}}$

Upon simplifying this gives:

$\begin{gathered} \frac{\lim_{x\to-\infty}\sqrt{\frac{4x^2}{x^2}-\frac{5}{x^2}}}{\lim_{n\to\infty}(1-\frac{3}{x})} \\ =\frac{\operatorname{\lim}_{x\to-\infty}\sqrt{4-\frac{5}{x^2}}}{\operatorname{\lim}_{n\to\infty}(1-\frac{3}{x})} \end{gathered}$

Taking the limit gives:

$\begin{gathered} \frac{\sqrt{4+\frac{5}{-\infty}}}{1-\frac{3}{-\infty}} \\ \frac{\sqrt{4+0}}{1-0} \\ =\frac{\pm2}{1} \\ =\pm2 \end{gathered}$