Limits of Continuous Functions

Key Questions

the limit is the value of the continuous function at the point concerned

Explanation:

the limit is the value of the continuous function at the point concerned

so for example

${\lim}_{x \to c} {x}^{2} = {c}^{2}$

• If the function is continuous at that value of $x$, just computing it should work. For other functions, try to factor the discontinuity out, and then computing it.

If nothing works, and you don't have access to derivatives to use L'HÃ´pital rule, just take a numeric approach and see if it's close to a pattern, or try to use other approaches like the Squeeze Theorem or a geomtrical proof. For example:

${\lim}_{x \rightarrow 4} \left({x}^{2} - 8 x + 16\right) = 0$

The function is continuous* for all real values of $x$ so just plugging the value of $x$ is good enough.

${\lim}_{x \rightarrow 4} \frac{{x}^{2} - 16}{x - 4} = {\lim}_{x \rightarrow 4} \frac{\left(x + 4\right) \left(x - 4\right)}{x - 4} = {\lim}_{x \rightarrow 4} \left(x + 4\right) = 8$

The function isn't continous at $x = 4$ because we'd have division by $0$, but using algebra we can take out the denominator, making a continuous function to just plug the number in.

${\lim}_{\theta \rightarrow 0} \frac{\sin \left(\theta\right)}{\theta} = 1$

This one you can't factor out. You need to use the unit circle and the Squeeze theorem to make a proof or try a numeric approach.

*A function is continuous on a certain range if you can draw the graph on that range without lifting your writing utensil, or more formally, if there isn't any point that would create a math error, like division by zero, even root of a negative, logarithm of a negative or null number, etc.

A continuous function is a function that is continuous at every point in its domain.

That is $f : A \to B$ is continuous if $\forall a \in A , {\lim}_{x \to a} f \left(x\right) = f \left(a\right)$

Explanation:

We normally describe a continuous function as one whose graph can be drawn without any jumps. That's a good place to start, but is misleading.

An example of a well behaved continuous function would be $f \left(x\right) = {x}^{3} - x$

graph{x^3-x [-2.5, 2.5, -1.25, 1.25]}

In fact any polynomial is well defined everywhere and continuous.

A less obvious example of a continuous function is $f \left(x\right) = \tan \left(x\right)$

graph{tan(x) [-10, 10, -5, 5]}

This appears to be discontinuous, with 'jumps' at $x = \frac{\pi}{2} + n \pi$ but those values of $x$ are excluded from the domain.

Similarly, the following function is continuous on its domain $\left(- \infty , 0\right) \cup \left(0 , \infty\right)$

$f \left(x\right) = \frac{x}{\left\mid x \right\mid}$

graph{x/abs(x) [-5, 5, -2.5, 2.5]}

If we add a definition of $f \left(0\right)$ then this becomes a discontinuous function.

$f \left(x\right) = \left\{\begin{matrix}\frac{x}{\left\mid x \right\mid} & \text{if x != 0" \\ 0 & "if x = 0}\end{matrix}\right.$

graph{(y-x/abs(x))(x^2+y^2-0.002) = 0 [-5, 5, -2.5, 2.5]}

This $f \left(x\right)$ fails the condition for continuity at the point $x = 0$.

${\lim}_{x \to 0} f \left(x\right)$ is not defined, let alone equal to $f \left(0\right)$