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Consider the continuity of (tan x)/x at the point x=0

by RoRi

Consider the function

    \[ f(x) = \frac{\tan x}{x} \qquad x \neq 0. \]

Determine what happens to f(x) as x \to 0. Define f(0) so that f will be continuous at x = 0, if possible.

First, we sketch the graph of f(x) = \frac{\tan x}{x} on the intervals \left[ -\frac{1}{4} \pi, 0\right) and \left( 0, \frac{1}{4} \pi \right].

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To determine what happens to f(x) as x \to 0 we can take the limit, (recalling that \lim_{x \to 0} \frac{\sin x}{x} = 1),

    \begin{align*} \lim_{x \to 0} \frac{\tan x}{x} &= \lim_{x \to 0} \frac{\sin x}{x \cos x} \\ & =\lim_{x \to 0} \left( \left( \frac{\sin x}{x} \right) \cos x \right) \\ & =\lim_{x \to 0} \cos x \\ &= 1. \end{align*}

Since f(x) = \frac{\tan x}{x} is not defined at x = 0 it is not continuous there. However, since the limit exists we could redefine f by

    \[ f(x) = \begin{cases} \frac{\tan x}{x} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0. \end{cases} \]

Then, this new f (which has the same values as the original function everywhere the original function is defined, but has the additional feature of being defined at x=0) is continuous at x=0 since it is defined there and \lim_{x \to 0} f(x) = f(0) = 1.

Note what we have done here. We had a function f(x) = \frac{\tan x}{x} which was undefined at the point x = 0. We created a new function (which we also called f, which is a bit confusing) that took the same values as the original function for all x \neq 0, and took the value 1 when x = 0. In this way we “removed” the discontinuity of the original f at the point x = 0, but we should be aware that this is actually a new function (since it has a different domain than the original).

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