Question

evaluate the following limits. if needed, enter oo for \\( \infty \\) and -oo for \\( -\infty \\).
(a) \\( \lim _{x \
ightarrow \infty} \frac{\sqrt{8+5 x^{2}}}{6+3 x}= \\)
(b) \\( \lim _{x \
ightarrow-\infty} \frac{\sqrt{8+5 x^{2}}}{6+3 x}= \\)
question help: video message instructor
submit question jump to answer

Explanation:

Step1: Divide numerator and denominator by \(x\)

For \(x\to\infty\), \(\lim_{x\to\infty}\frac{\sqrt{8 + 5x^{2}}}{6+3x}=\lim_{x\to\infty}\frac{\sqrt{\frac{8}{x^{2}}+5}}{\frac{6}{x}+3}\)
Since \(\lim_{x\to\infty}\frac{8}{x^{2}} = 0\) and \(\lim_{x\to\infty}\frac{6}{x}=0\)

Step2: Calculate the limit

\(\lim_{x\to\infty}\frac{\sqrt{\frac{8}{x^{2}}+5}}{\frac{6}{x}+3}=\frac{\sqrt{0 + 5}}{0+3}=\frac{\sqrt{5}}{3}\)

For \(x\to-\infty\), \(\lim_{x\to-\infty}\frac{\sqrt{8 + 5x^{2}}}{6+3x}=\lim_{x\to-\infty}\frac{\sqrt{\frac{8}{x^{2}}+5}}{\frac{6}{x}+3}\)
But when \(x\to-\infty\), \(\sqrt{x^{2}}=-x\) (because \(x<0\)), so \(\lim_{x\to-\infty}\frac{\sqrt{8 + 5x^{2}}}{6+3x}=\lim_{x\to-\infty}\frac{\sqrt{\frac{8}{x^{2}}+5}}{\frac{6}{x}+3}\times\frac{-1}{-1}=\lim_{x\to-\infty}\frac{-\sqrt{\frac{8}{x^{2}}+5}}{\frac{6}{x}+3}\)
Since \(\lim_{x\to-\infty}\frac{8}{x^{2}} = 0\) and \(\lim_{x\to-\infty}\frac{6}{x}=0\)
\(\lim_{x\to-\infty}\frac{-\sqrt{\frac{8}{x^{2}}+5}}{\frac{6}{x}+3}=\frac{-\sqrt{0 + 5}}{0+3}=-\frac{\sqrt{5}}{3}\)

Answer:

(a) \(\frac{\sqrt{5}}{3}\)
(b) \(-\frac{\sqrt{5}}{3}\)