Question

evaluate the following limits. if needed, enter oo for $infty$ and -oo for $-infty$. (a) $lim_{x
ightarrowinfty}\frac{sqrt{4 + 11x^{2}}}{7 + 2x}=$ (b) $lim_{x
ightarrow-infty}\frac{sqrt{4 + 11x^{2}}}{7 + 2x}=$ question help: video

Explanation:

Step1: Divide numerator and denominator by $x$

For $x\to\infty$ and $x\to -\infty$, we divide $\frac{\sqrt{4 + 11x^{2}}}{7+2x}$ by $x$. When $x
eq0$, we can write $\frac{\sqrt{4 + 11x^{2}}}{7 + 2x}=\frac{\sqrt{\frac{4}{x^{2}}+11}}{\frac{7}{x}+2}$ (since $\sqrt{x^{2}}=|x|$). When $x\to\infty$, $|x| = x$ and when $x\to-\infty$, $|x|=-x$.

Step2: Evaluate limits of individual terms

We know that $\lim_{x\to\pm\infty}\frac{4}{x^{2}} = 0$ and $\lim_{x\to\pm\infty}\frac{7}{x}=0$.

Step3: Find limit as $x\to\infty$

For $\lim_{x\to\infty}\frac{\sqrt{\frac{4}{x^{2}}+11}}{\frac{7}{x}+2}$, substituting the limits of the terms, we get $\frac{\sqrt{0 + 11}}{0+2}=\frac{\sqrt{11}}{2}$.

Step4: Find limit as $x\to-\infty$

For $\lim_{x\to-\infty}\frac{\sqrt{4 + 11x^{2}}}{7 + 2x}$, we first note that when dividing by $x$ (where $x<0$), $\frac{\sqrt{4 + 11x^{2}}}{x}=-\sqrt{\frac{4}{x^{2}}+11}$. So $\lim_{x\to-\infty}\frac{\sqrt{4 + 11x^{2}}}{7 + 2x}=\lim_{x\to-\infty}\frac{-\sqrt{\frac{4}{x^{2}}+11}}{\frac{7}{x}+2}=-\frac{\sqrt{11}}{2}$.

Answer:

(a) $\frac{\sqrt{11}}{2}$
(b) $-\frac{\sqrt{11}}{2}$