Answer
$$
\lim\limits_{x \to \infty} \frac{\sqrt {36x^{2}+2x+7}}{3x}
$$
(a)
A graphing calculator can give a deceptive view of a function. The Figure shows the graph appears to have horizontal asymptotes at $y=\pm 2 $. Thus we determine that :
$$
\lim\limits_{x \to \infty} \frac{\sqrt {36x^{2}+2x+7}}{3x}=2
$$
(b)
As $x \rightarrow \infty$
$$
\frac{\sqrt {36x^{2}+2x+7}}{3x} \rightarrow \frac{6|x|}{3 x}
$$
since $x \gt 0,|x|= x,$ so
$$
\frac{6 |x|}{3 x}=\frac{6 x}{3 x}=\frac{6}{2}=2
$$
and, to evaluate the limit at infinity of a rational function, divide the numerator and denominator by the largest power of the variable that appears in the denominator, $x$ here, and then use these results. Thus, we find that
\[
\lim\limits_{x \to \infty} \frac{\sqrt {36x^{2}+2x+7}}{3x}=2
\]
Work Step by Step
$$
\lim\limits_{x \to \infty} \frac{\sqrt {36x^{2}+2x+7}}{3x}
$$
(a)
A graphing calculator can give a deceptive view of a function. The Figure shows the graph appears to have horizontal asymptotes at $y=\pm 2 $. Thus we determine that :
$$
\lim\limits_{x \to \infty} \frac{\sqrt {36x^{2}+2x+7}}{3x}=2
$$
(b)
As $x \rightarrow \infty$
$$
\frac{\sqrt {36x^{2}+2x+7}}{3x} \rightarrow \frac{6|x|}{3 x}
$$
since $x \gt 0,|x|= x,$ so
$$
\frac{6 |x|}{3 x}=\frac{6 x}{3 x}=\frac{6}{2}=2
$$
and, to evaluate the limit at infinity of a rational function, divide the numerator and denominator by the largest power of the variable that appears in the denominator, $x$ here, and then use these results. Thus, we find that
\[
\lim\limits_{x \to \infty} \frac{\sqrt {36x^{2}+2x+7}}{3x}=2
\]
