12. Limits
A function requires an input to get an output. As an input gets closer and closer to a particular number, so too do the corresponding outputs get closer and closer to a particular value. Finding this value that the output is approaching is finding the limit of the function. This is a fundamental mathematical concept that is especially important in calculus.
Do you need this chapter?
Below is a quiz on limits. If you can answer all these questions, then you can skip this chapter as you already have the requisite knowledge.
Quiz: Limits
Determine the following limits:
1. [latex]\begin{align*}\lim_{x\rightarrow0}{4x+2}\end{align*}[/latex]
2. [latex]\begin{align*}\lim_{x\rightarrow2}\dfrac{9-x^{2}}{4}\end{align*}[/latex]
3. [latex]\begin{align*}\lim_{x\rightarrow0}2x+h\end{align*}[/latex]
4. [latex]\begin{align*}\lim_{x\rightarrow3}\dfrac{x^{2}-9}{x-3}\end{align*}[/latex]
5. [latex]\begin{align*}\lim_{h\rightarrow0}\dfrac{2xh+h^{2}}{h}\end{align*}[/latex]
Answers
1. [latex]2[/latex]
2. [latex]\dfrac{5}{4}[/latex]
3. [latex]h[/latex]
4. [latex]6[/latex]
5. [latex]2x[/latex]
Suppose that [latex]f(x)=x-2[/latex]. If we let [latex]x=2[/latex] we get [latex]f(2)=2-2=0[/latex].
Similarly, we could say that the limit of the function, [latex]f(x)=x-2[/latex], as [latex]x[/latex] approaches [latex]2[/latex], is equal to zero, that is,
\begin{align*}
\lim_{x\rightarrow2}\left(x-2\right)&=0
\end{align*}
In other words, as the value of [latex]x[/latex] approaches [latex]2[/latex], either from above or from below, the value of [latex]f(x)[/latex], will approach zero.
We can assume that the limit of a function as [latex]x[/latex] approaches a particular value is equal to the value of the function - if the function exists at that point.
The use of limits, however, is particularly helpful when [latex]f(x)[/latex] does not exist for a certain value of [latex]x[/latex]. For example, the function
\begin{align*}
f(x) &= \dfrac{x-2}{x^2-4}
\end{align*}
does not exist at [latex]x=2[/latex] because
\begin{align*}f(2)&=\dfrac{2-2}{2^2-4}\\
&=\dfrac{0}{0}\end{align*}
which is undefined.
Even though this function is undefined when [latex]x[/latex] is equal to [latex]2[/latex], we can still find the limit of the function as [latex]x[/latex] approaches [latex]2[/latex].
That is,
\begin{align*}
\lim_{x\rightarrow2}\frac{x-2}{x^2-4}
&=\lim_{x\rightarrow2}\frac{x-2}{(x-2)(x+2)}\\
&=\lim_{x\rightarrow2}\frac{1}{x+2} \\
&=\frac{1}{2+2} \\
&=\frac{1}{4}.
\end{align*}
We can check this answer by putting a value very close to [latex]2[/latex] into the original function and we should get an answer very close to [latex]\dfrac{1}{4}[/latex]. If we let x=1.999, we have
\begin{align*}
f \left(1.999 \right) &= \dfrac{1.999-2}{1.999^2-4}\\
& \approx 0.25006
\end{align*}
which is close to [latex]\dfrac{1}{4}[/latex] as expected.
Examples: Limits
Determine the following limits:
1. \begin{align*}\lim_{x\rightarrow0}{(7x^2+2x)}&=7(0)^2+2(0)\\
&=0.\end{align*}
2. \begin{align*}\lim_{x\rightarrow8}{(9-x)}&=9-8\\
&=1.\end{align*}
3. \begin{align*}\lim_{x\rightarrow5}\dfrac{x^2-25}{x-5}
&=\lim_{x\rightarrow5}\dfrac{(x-5)(x+5)}{(x-5)}\\
&=\lim_{x\rightarrow5}{x+5} \\
&=10.\end{align*}
4. \begin{align*}\lim_{h\rightarrow0}\dfrac{2xh+h^{2}}{h}&=\lim_{h\rightarrow0}\dfrac{h(2x+h)}{h}\\
&=\lim_{h\rightarrow0}{2x+h} \\
&=2x.\end{align*}
5. \begin{align*}\lim_{h\rightarrow0}\dfrac{(x+h)^2-x^2}{h}&=\lim_{h\rightarrow0}\frac{x^{2}+ 2xh+ h^{2}-x^{2}}{h}\\
&=\lim_{h\rightarrow0}\frac{2xh+ h^{2}}{h} \\
&=\lim_{h\rightarrow0}\frac{h(2x+ h)}{h} \\
&=\lim_{h\rightarrow0}2x +h\\
&=2x.\end{align*}
6. \begin{align*}\lim_{h\rightarrow0}\dfrac{(x+h)^3-x^3}{h}
&=\lim_{h\rightarrow0}\frac{x^{3}+3x^2h+3xh^2+ h^{3}-x^{3}}{h}\\
&=\lim_{h\rightarrow0}\frac{3x^2h+3xh^2+h^{3}}{h}\\
&=\lim_{h\rightarrow0}\frac{h(3x^2+3xh+ h^2)}{h}\\
&=\lim_{h\rightarrow0}(3x^2+3xh+h^2)\\
&=3x^{2}.\end{align*}
Note: if trying to evaluate the limit gives the indeterminate form [latex]\dfrac{0}{0}[/latex], then the expression must be simplified first before finding the limit. This frequently involves algebraic manipulation of the numerator, the denominator, or both.
Now try the questions in the exercise below.
Exercises: Limits
Determine the following limits:
1. [latex]\begin{align*}\lim_{x\rightarrow0}{\left(x^2-2\right)}.\end{align*}[/latex]
2. [latex]\begin{align*}\lim_{x\rightarrow1}{\left(1-x\right)}.\end{align*}[/latex]
3. [latex]\begin{align*}\lim_{x\rightarrow1}\dfrac{x-1}{x^2-1}.\end{align*}[/latex]
4. [latex]\begin{align*}\lim_{x\rightarrow3}\dfrac{9-x^2}{3-x}.\end{align*}[/latex]
5. [latex]\begin{align*}\lim_{h\rightarrow0}\dfrac{h\cos\left(h\right)}{h}.\end{align*}[/latex]
6. [latex]\begin{align*}\lim_{h\rightarrow0}\dfrac{2\left(x+h\right)^2-2x^2}{h}.\end{align*}[/latex]
7. [latex]\begin{align*}\lim_{h\rightarrow0}\dfrac{\left(x+h\right)^2+2\left(x+h\right)+1-\left(x^2+2x+1\right)}{h}.\end{align*}[/latex]
8. [latex]\begin{align*}\lim_{h\rightarrow0}\dfrac{5\left(x+h\right)^3-5x^3}{h}.\end{align*}[/latex]
Answers
1. [latex]-2[/latex]
2. [latex]0[/latex]
3. [latex]\dfrac{1}{2}[/latex]
4. [latex]6[/latex]
5. [latex]1[/latex]
6. [latex]4x[/latex]
7. [latex]2x+2[/latex]
8. [latex]15x^2[/latex]
Key takeaways
- A limit describes the value that a function approaches as the input gets closer and closer to a particular number. It is a fundamental mathematical concept that is especially important in calculus.
- We can assume that the limit of a function as [latex]x[/latex] approaches a particular value is equal to the value of the function - if the function exists at that point.
- The use of limits is particularly helpful when [latex]f(x)[/latex] does not exist for a certain value of [latex]x[/latex].
- If trying to evaluate the limit gives the indeterminate form [latex]\dfrac{0}{0}[/latex], then the expression must be simplified first before finding the limit.
