Also known as end behavior, here we’ll look at what functions do as $x$ approaches infinity. For some functions, infinite $x$ means infinite $f(x)$.
\[\lim_{x\to\infty}x^2 = \infty \qquad \lim_{x\to -\infty}x^2 = \infty\]But in other cases, you can run into horizontal asymptotes. You might remember these from Algebra 2, and that there were some tricks to spotting these without going though the trouble of graphing or substituting values, though these only work with rational functions.
The first case is when the degree of the denominator is more than the degree of the numerator. In this case, the limits as $x$ approaches $\pm\infty$ will be 0. Since the denominator grows much quicker than the numerator, you’ll be dividing by larger and larger numbers, meaning closer and closer to 0.
The opposite case means the graph will take off to infinity, with the sign determining which way.
In the case where they are equal, the coefficients of the leading terms determines the limit. The limit of $\frac{3x^2-1}{2x^2}$ will be $\frac{3}{2}$.
When you’re not dealing with rational functions, you can go about it the old fashioned way with graphs and tables. Knowing your limit properties will also help.
Be mindful that you can have two horizontal asymptotes, where the end of the function will behave differently. The logistic function is one case.
\[f(x)=\frac{1}{1+e^{-x}}\]Make sure to test with graphs and values in cases you’re not sure of.
To find $\lim_{x\to\infty}\frac{\sin x}{x}$ we can use the Squeeze Theorem. We can start with the knowledge that $\sin x$ is between $-1$ and $1$, then modify the inequality to arrive at our desired function.
\[-1 \le \sin x \le 1 \\[0.5em] -\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}\]Since we know $\lim_{x\to\infty}\frac{1}{x}=0$, as does its negative counterpart, then $\lim_{x\to\infty}\frac{\sin x}{x}=0$ as well.