# Determining When a Limit does not Exist

## Key Questions

• Remember that limits represent the tendency of a function, so limits do not exist if we cannot determine the tendency of the function to a single point. Graphically, limits do not exist when:

1. there is a jump discontinuity
(Left-Hand Limit $\ne$ Right-Hand Limit)
The limit does not exist at $x = 1$ in the graph below.

1. there is a vertical asymptote
(Infinit Limit)
(Caution: When you have infinite limits, limits do not exist.)
The limit at $x = 2$ does not exist in the graph below.

1. there is a violent oscillation
(e.g., $\sin \left(\frac{1}{x}\right)$ at $x = 0$, shown below)

I hope that this was helpful.

• =$\infty$

Solution

=lim_(x→4)((x-4)/(x^2-8x+16)), plugging the limit we get $\frac{0}{0}$

=lim_(x→4)((x-4)/(x^2-4x-4x+16))

=lim_(x→4)((x-4)/(x(x-4)-4(x-4)))

=lim_(x→4)(x-4)/((x-4)^2)

=lim_(x→4)1/((x-4))

Now applying the limit, we get

$= \frac{1}{0} = \infty$, which implies limit does not exist

• In short, the limit does not exist if there is a lack of continuity in the neighbourhood about the value of interest.

Recall that there doesn't need to be continuity at the value of interest, just the neighbourhood is required.

Most limits DNE when ${\lim}_{x \to {a}^{-}} f \left(x\right) \ne {\lim}_{x \to {a}^{+}} f \left(x\right)$, that is, the left-side limit does not match the right-side limit. This typically occurs in piecewise or step functions (such as round, floor, and ceiling).

A common misunderstanding is that limits DNE when there is a point discontinuity in rational functions. On the contrary, the limit exists perfectly at the point of discontinuity!

So, an example of a function that doesn't have any limits anywhere is f(x) = {x=1, x in QQ; x=0, otherwise}. This function is not continuous because we can always find an irrational number between 2 rational numbers and vice versa.