3.6.1 Limits

Usually the limits are the simplest kind where simple substitution can be used to get an appropriate answer. For example:

$$\lim_{x \to 3} 3x^2 - 4 = 3 \cdot 3^2 - 4 = 23$$

However certain problems, which when passing the limit, might lead to a $0/0$ violation. In this case, you want to see if there are any common factors that you can cancel so that passing the limit doesn’t give you an indeterminate form. Let’s look at an example:

$$\lim_{x \to 2} \frac{(x - 2)(x + 3)}{(x + 5)(x - 2)} = \lim_{x \to 2} \frac{(x + 3)}{(x + 5)} = \frac{5}{7}$$

If we had plugged in $x = 2$ into the original problem, we would have gotten a $0/0$ form, however after canceling the factors, we were able to pass the limit.

The final testable question concerning limits involve l’hˆopitals rule (this requires the understanding of derivatives in order to compute it, see the next section for instructions on how to compute that). L’hˆopitals rule states that if you come across a limit that gives an indeterminant form (either $0/0$ or $\infty/\infty$) you can compute the limit by taking the derivative of both the numerator and the denominator then passing the limit. So:

$$\lim_{x \to n} \frac{f(x)}{g(x)} = \frac{0}{0} \text{ or } \frac{\infty}{\infty} \Rightarrow \lim_{x \to n} \frac{f(x)}{g(x)} = \lim_{x \to n} \frac{f'(x)}{g'(x)} \Rightarrow \frac{f'(n)}{g'(n)}$$

Let’s look at an example of l’hˆopitals rule with computing the limit $\lim_{x \to 0} \frac{\sin x}{x}$:

$$\lim_{x \to 0} \frac{\sin x}{x} = \frac{0}{0} \Rightarrow \lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{(\sin x)'}{x'} = \lim_{x \to 0} \frac{\cos x}{1} = 1$$

The following are some more practice problems with limits:

Problem Set 3.6.1