One-sided limits
Many real-life quantities change formula at a certain point: a tariff that goes up after a certain number of minutes, a tax that changes bracket, a speed that drops suddenly. To study the limit at that point of change, you have to look at each side separately, and only then decide whether a two-sided limit exists.
One-sided limits: definition
A one-sided limit only takes into account the values of $x$ that approach $a$ from a single side.
Given the function $f(x) = x^2 - 3$, we want to investigate what happens when the values of $x$ get very close to $2$. Complete the following table of values and, in…
Given the function $f(x) = x^2 - 3$, we want to investigate what happens when the values of $x$ get very close to $2$. Complete the following table of values and, in light of the results, state what you think the value of $\lim_{x \to 2} f(x)$ is.
| $x$ | $1,9$ | $1,99$ | $1,999$ | $2,001$ | $2,01$ | $2,1$ |
|---|---|---|---|---|---|---|
| $f(x)$ |
The existence criterion: when there is a two-sided limit
The two one-sided limits on their own guarantee nothing about the two-sided limit: you have to compare them.
Observe the following graph of the function $g(x)$ and determine the value of the one-sided limits at the boundary point $x = 1$. Also state whether the overall limit…
Observe the following graph of the function $g(x)$ and determine the value of the one-sided limits at the boundary point $x = 1$. Also state whether the overall limit $\lim_{x \to 1} g(x)$ exists.
- a) $\lim_{x \to 1^-} g(x)$
- b) $\lim_{x \to 1^+} g(x)$
Given the function $f(x) = \frac{x^2 - 1}{x^2 - 2x + 1}$, calculate the one-sided limits at the point where the denominator vanishes. a) What happens to the values of…
Given the function $f(x) = \frac{x^2 - 1}{x^2 - 2x + 1}$, calculate the one-sided limits at the point where the denominator vanishes.
- a) What happens to the values of the function as we approach $x = 1$ from the left?
- b) And from the right?
- c) Based on your results, can we say that the overall limit at $x = 1$ exists?
Hint. Factor the numerator and the denominator and simplify the fraction before studying the sign on each side.
Piecewise functions: finding parameters so that the limit exists
The most common exercise in this topic in the university entrance exam (EvAU): given a parameter in a piecewise function, find its value so that the limit exists at the joining point.
Determine the value of the parameter $k$ so that the limit of the following function exists at its boundary point $x = 2$:…
Determine the value of the parameter $k$ so that the limit of the following function exists at its boundary point $x = 2$:
$f(x) = \begin{cases} kx - 3 & \text{if } x < 2 \\ x^2 + k & \text{if } x \geq 2 \end{cases}$
For the limit to exist, remember that the one-sided limits must be equal. Once $k$ is found, state the value of the limit.
Let the function be $h(x) = \frac{\sqrt{x^2 + k} - 3}{x-1}$. a) Determine the value of $k \in \mathbb{R}$ such that the function gives an indeterminate form of the type…
Let the function be $h(x) = \frac{\sqrt{x^2 + k} - 3}{x-1}$.
- a) Determine the value of $k \in \mathbb{R}$ such that the function gives an indeterminate form of the type $0/0$ when $x \to 1$.
- b) For this value of $k$, calculate the one-sided limits at $x=1$ and investigate whether the overall limit exists.
Absolute value as a piecewise function
A function with absolute value can always be rewritten without it, by splitting into the intervals where the argument is positive or negative. Once this is done, the study of the one-sided limits is exactly the same.
Calculate the one-sided limits at the break points of the following function, which combines a modulus with a rational function: $$g(x) = \frac{x^2 - 5x + 6}{|x-3| \cdot…
Calculate the one-sided limits at the break points of the following function, which combines a modulus with a rational function:
$$g(x) = \frac{x^2 - 5x + 6}{|x-3| \cdot (x-1)}$$From the results obtained, reason whether $\lim_{x \to 3} g(x)$ exists and describe the behaviour of the function near $x=3$ and $x=1$.