Question

evaluate the following limits. if needed, enter oo for \\(\infty\\) and -oo \\(\text{(a)} \lim_{x \to \infty} \frac{\sqrt{6 + 10x^2}}{5 + 7x} = \\) \\(\text{(b)} \lim_{x \to -\infty} \frac{\sqrt{6 + 10x^2}}{5 + 7x} = \\) question help: video message instructor

Explanation:

Step1: For part (a), divide numerator and denominator by \(x\) (since \(x \to \infty\), \(x>0\), so \(\sqrt{x^2}=x\))

First, rewrite the numerator and denominator by dividing each term by \(x\):
The numerator \(\sqrt{6 + 10x^2}\) divided by \(x\) is \(\frac{\sqrt{6 + 10x^2}}{x}=\sqrt{\frac{6 + 10x^2}{x^2}}=\sqrt{\frac{6}{x^2}+10}\) (because \(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\) for \(a\geq0,b>0\) and here \(x>0\) so \(x = \sqrt{x^2}\)).
The denominator \(\frac{5 + 7x}{x}=\frac{5}{x}+7\).
So the limit becomes \(\lim_{x\to\infty}\frac{\sqrt{\frac{6}{x^2}+10}}{\frac{5}{x}+7}\).

Step2: Evaluate the limit as \(x\to\infty\)

As \(x\to\infty\), \(\frac{6}{x^2}\to0\) and \(\frac{5}{x}\to0\).
So substituting these into the expression, we get \(\frac{\sqrt{0 + 10}}{0 + 7}=\frac{\sqrt{10}}{7}\).

Step3: For part (b), divide numerator and denominator by \(x\) (since \(x \to -\infty\), \(x<0\), so \(\sqrt{x^2}=-x\))

Rewrite the numerator and denominator by dividing each term by \(x\) (note that \(x\) is negative here, so \(\frac{\sqrt{6 + 10x^2}}{x}=\frac{\sqrt{6 + 10x^2}}{-\sqrt{x^2}}=-\sqrt{\frac{6 + 10x^2}{x^2}}=-\sqrt{\frac{6}{x^2}+10}\) because \(x = -\sqrt{x^2}\) when \(x<0\)).
The denominator \(\frac{5 + 7x}{x}=\frac{5}{x}+7\).
So the limit becomes \(\lim_{x\to-\infty}\frac{-\sqrt{\frac{6}{x^2}+10}}{\frac{5}{x}+7}\).

Step4: Evaluate the limit as \(x\to-\infty\)

As \(x\to-\infty\), \(\frac{6}{x^2}\to0\) and \(\frac{5}{x}\to0\).
Substituting these into the expression, we get \(\frac{-\sqrt{0 + 10}}{0 + 7}=-\frac{\sqrt{10}}{7}\).

Answer:

(a) \(\frac{\sqrt{10}}{7}\)
(b) \(-\frac{\sqrt{10}}{7}\)