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Calculus Notes

Section 1.3: Evaluating Limits Analytically

For "normal" functions like $f(x) = 2x$, calculating limits is straightforward. All one needs to do is plug the appropriate value of $x$ into the function. For example, one can show that $\lim_{x\to 3} 2 x = 6$ just by plugging $ x= 3$ into the function $f(x) = 2x$.

However, there are other limits for which one cannot simply plug the value $x$ into the function, because the function does not exist at the particular $x$-value, or because the value of the function at that point and the value it is approaching are different. For the purpose of calculus, it is these special cases that we are most interested in.

Often there is some algebraic "trick" one can use to evaluate limits. For example, the limit $$\lim_{x\to 1} \frac{x^3-1}{x-1}$$ may be evaluated by factoring the numerator and canceling out the denominator. Once the problematic $x-1$ term has been canceled out, it is possible to plug in $x=1$ into the remainder. \begin{align*} \lim_{x\to 1} \frac{x^3-1}{x-1} &= \lim_{x\to 1} \frac{(x-1)(x^2+x+1)}{x-1}\\ &= \lim_{x\to 1} (x^2 + x+ 1) \\ &= 3 \end{align*}

The fact that the original function did not exist at $x=1$ is immaterial since the limit is only concerned with the function's behavior as it approaches $x=1$.

Another useful trick is "multiplying by the conjugate." For example, \begin{align*} \lim_{x\to 3} \frac{\sqrt{x+1} -2}{x-3} &= \lim_{x\to 3} \frac{\sqrt{x+1} -2}{x-3} \cdot \frac{\sqrt{x+1} +2}{\sqrt{x+1} +2}\\ &=\lim_{x\to 3} \frac{x-3}{(x-3)(\sqrt{x+1}+2)}\\ &=\lim_{x\to 3} \frac{1}{\sqrt{x+1}+2}\\ &=\frac{1}{4} \end{align*}

Squeeze Theorem

The squeeze theorem states that if one function always lies between two functions over an open interval containing $c$, and that the limit of both outside functions at $c$ is $L$, then the limit of the middle function must be $L$ also (see Figure 1).

The squeeze theorem can be used to prove two trigonometric limits of particular importance:

$$\lim_{x\to 0} \frac{\sin x}{x} = 1 \hspace{15pt} \text{and} \hspace{15pt} \lim_{x\to 0} \frac{1 - \cos x}{x} = 0$$