Computing Limits

Simplifying a Problem Algebraically

$$\lim_{x\to 7}\frac{t^2-5t-14}{t-7}$$

$$\lim_{x\to 1}\frac{x^2-3x+2}{x^2-1}$$

$$\lim_{x\to 2}\frac{2-x}{2-\sqrt{x+2}}$$

$$\lim_{h\to 0}\frac{(h-1)^2-1}{h}$$

Some Useful Limits:

$$\lim_{t\to\pm\infty}\frac{1}{t}=0$$

In fact, of $r>0$ then $$\lim_{t\to\infty}\frac{1}{t^r}=0$$

$$\lim_{t\to\infty}e^x=\infty$$

$$\lim_{t\to-\infty}e^x=0$$

The Limit Laws

Sum, difference scalar multiple, product, quotient

$$\lim_{x\to a}(f(x)+g(x))=\lim_{x\to a}f(x)+\lim_{x\to a}g(x)$$

$$\lim_{x\to a}(f(x)-g(x))=\lim_{x\to a}f(x)-\lim_{x\to a}g(x)$$

$$\lim_{x\to a}(c f(x))=c\lim_{x\to a}f(x)$$

$$\lim_{x\to a}\Big(f(x)g(x)\Big)=\Big(\lim_{x\to a}f(x)\Big)\Big(\lim_{x\to a}g(x)\Big)$$

$$\lim_{x\to a}\Big(\frac{f(x)}{g(x)}\Big)=\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}$$

The same laws hold, even when the point $a$ is taken to be infinity

Computing Limits At Infinity

Compute the limits: $$\lim_{x\to\infty}\frac{3x^5-2x^2}{5x^7+1}$$ $$\lim_{x\to\infty}\frac{3x^5-2x^2}{x^5-4}$$

Find the horiontal asymptotes of the function $$f(x)=\frac{\sqrt{2x^2+1}}{3x-5}$$

The Squeeze Theorem

Sometimes it’s hard to compute a limit directly, because it has some complicated functions involved for which our above rules don’t work the best. For example, it’s hard to immediately figure out the end-behavior of $2\sin(x^3)/x$, because of the $\sin$ function. In certain cases like this, there’s a nice trick that can be used: instead of attacking the problem directly we instead replace this one difficult problems with two easier ones: we create a function $g(x)$ which is less than our mystery function to serve as a lower estimate, and we also create a function $f(x)$ to serve as an upper estimate. Then, we may try to understand the limiting behavior of our original function in terms of the limiting behavior of $f$ and $g$, effectively squeezing our function between them.

The Squeeze Theorem If $h$ is a function whose limit we wish to calculate, and we can find functions $f,g$ such that for all $x$ $g(x)\leq h(x)\leq f(x)$, whose limits we can calculate, and $$\lim_{x\to a}g(x)=\lim_{x\to a}f(x)$$ **then** also $\lim_{x\to a}h(x)$ exists, and is equal to this same value.

The most useful examples for us will spring forth from the fact that we can easily bound $\sin$ and $\cos$ by constants: for every single $x$ the values of these trigonometric functions lie in $[-1,1]$. Thus, given a function like $2\sin(x^3)/x$, we can bound this above and below using $-1$ and $1$:

$$\frac{2(-1)}{x}\leq \frac{2\sin x^3}{x}\leq \frac{2(1)}{x}$$

The two bounds are functions for which its easy to take the limit, and both have limit $0$ at infinity! Thus, $$\lim_{x\to\infty}\frac{2\sin(x^3)}{x}=0$$