MA101.10 How to find limits

If a function is know to be  continuous at x=a, then by the definition of continuity we have, \displaystyle\lim_{t\to a} f(x) = f(a).

We will take is as known that the elementary functions – polynomial, rational (polynomial/polynomial), trigonometrical, exponential, logarithmic) – are continuous everywhere.

Therefore, although not very exciting, we are able to say such things as:

\displaystyle\lim_{x\to \frac{\pi}{2}} sin x = sin \frac{\pi}{2}=1

\displaystyle\lim_{x\to 3} e^x = e^3

\displaystyle\lim_{x\to 2} \frac{x(x-1)}{x-3} = \frac{2(2-1)}{2-3} = -2

A very useful rule (it’s a theorem really) for finding limits is L’Hôpitals rule (LH), which is really a collection of rules. The most comprehensive form of the rule is as follows:

Suppose you wish to find something like \displaystyle\lim_{x\to[\;]} \frac{f(x)}{g(x)} where [ ] may be a, a^+, a^-, \infty, -\infty.

Suppose further that \displaystyle\lim_{x\to[\;]} f(x) = \lim_{x\to [\;]} g(x) = \{\} where either

\{\}=0  – (\frac{0}{0} type), or
\{\}=\infty – (\frac{\infty}{\infty} type).

Then the LH rule says look at \displaystyle\lim_{x\to [\;]} \frac{f'(x)}{g'(x)} and if you can do it, say find it to be ( ) then the original limit is the same.

Danger: Do not use LH when it is not applicable.

For example \displaystyle\lim_{x\to 0} \frac{sin(x) + 1}{sin(x)+2} = \frac{1}{2} is not the same as \displaystyle\lim_{x\to 0} \frac{cos(x)}{cos(x)} = 1.

The LH rule gives us with no effort some useful results. For example:

\displaystyle\lim_{x\to 0} \frac{sin(x)}{x} = \lim_{x\to 0} \frac{cos(x)}{1} = 1