This article is full of everyone's favorite things: examples! Want an example of finding a limit algebraically? It is here. Want an example of finding the limit of a rational function? That is here too! Even better, there are examples involving the exponential function and piecewise-defined functions.
Finding the Limit of Rational Functions
The limit of rational functions is the number at which a rational function gets closer \(f(x) \rightarrow b\) as \(x\) gets closer to a certain value \(a\).
\[lim_{x \rightarrow a} \dfrac{f(x)}{g(x)}=b\]
Remember that rational functions are continuous on their domains, so at any point in the domain of a rational function finding the limit is as easy as finding the function value at that point. It starts to get a bit more fun at points that aren't in the domain or in finding the limit at infinity.
Find
\[lim_{x \rightarrow 2} \dfrac{2x^2-3x+1}{x^3+4}\]
Answer:
The idea is to apply the Quotient Rule for limits if possible. Since the numerator and denominator are both polynomials.
\[lim_{x \rightarrow 2} (2x^2-3x+1)=2(2)^2-3(2+1\]
\[lim_{x \rightarrow 2} = 8-6+1\]
\[lim_{x \rightarrow 2}=3\]
and
\[lim_{x \rightarrow 2}(x^3+4)=2^3+4\]
\[lim_{x \rightarrow 2}(x^3+4)=12\]
which means the conditions to apply the Quotient Rule for limits is met. Now you know that:
\[lim_{x \rightarrow 2} \dfrac{2x^2-3x+1}{x^3+4}=\dfrac{3}{12}\]
\[lim_{x \rightarrow 2} = \dfrac{1}{4}\].
Now find
\[\lim_{x \rightarrow \infty} \dfrac{x^2+2x+4}{x^3-8}\].
Answer:
While the Quotient Rule holds for limits at infinity, it does require that the limit of the numerator and denominator are both real numbers, which in this case isn't true. That means you can't apply the Quotient Rule for limits at infinity. Instead, try factoring to see if that will help. If you factor the denominator, you will see that
\[x^3-8=(x-2)(x^2+2x+4)\]
Let's cancel some factors out! This leaves us with
\[lim_{x \rightarrow \infty } \dfrac{x^2+2x+4}{x^3-8}=lim_{x \rightarrow \infty} \dfrac{x^2+2x+4}{(x-2)(x^2+2x+4)}\]
\[lim_{x \rightarrow \infty } \dfrac{1}{x-2}\]
This is a much simpler limit; for more information on limits like this, see Infinite Limits. There you will learn how to show that
\[lim_{x \rightarrow \infty} \dfrac{1}{x-2}=0\].
That means
\[lim_{x \rightarrow \infty} \dfrac{x^2+2x+4}{x^3-8}=0\]
In the following example, you can see what happens when there is a vertical asymptote where you are trying to take the limit.
Find
\[lim_{ x \rightarrow 2} \dfrac{x^2+2x+4}{x^3-8}\].
Answer:
In the previous example, you were able to factor the denominator, enabling you to look at a simpler limit:
\[lim_{x \rightarrow 2} \dfrac{1}{x-2}\]
For this example, you will need to look at the limit from the left and the limit from the right and see if they are the same. In fact
\[lim_{x \rightarrow 2^+} \dfrac{1}{x-2}=\infty\]
while
\[lim_{x \rightarrow 2^-} \dfrac{1}{x-2}=-\infty\]
So, you can't find the limit, and you would say that the limit doesn't exist.
Finding the Limit of a Function Algebraically
There are lots of algebra techniques you can use to help you find limits. One of the most frequently used is simplifying fractions.
Find
\[\lim_{x \rightarrow 2} \dfrac{\dfrac{1}{x+1}-\dfrac{1}{3}}{x-2}\]
Answer:
Notice that this function has something interesting going on at \(x=2\) since the denominator is zero there. It will either be a hole in the graph, or a vertical asymptote, or some other discontinuity. That means you can't apply the Quotient Rule for limits since the limit of the denominator cannot be zero. Instead, let's do some algebra first:
\[\dfrac{\dfrac{1}{x+1}-\dfrac{1}{3}}{x-2} = \left( \dfrac{1}{x-2} \right) \left( \dfrac{1}{x+1}-\dfrac{1}{3} \right)\]
\[\left( \dfrac{1}{x-2} \right) \left( \dfrac{3-(x+1)}{3(x+1)} \right)\]
\[\left( \dfrac{1}{x-2} \right) \left( \dfrac{2-x}{3(x+1)} \right)\]
\[-\left( \dfrac{1}{x-2} \right) \left( \dfrac{x-2}{3(x+1)} \right)\]
\[-\dfrac{1}{3} \left( \dfrac{1}{x+1} \right)\]
Now you can use the
limit laws to see that\[lim_{x \rightarrow 2} \left( - \dfrac{1}{3} \left( \dfrac{1}{x+1} \right) \right)=-\dfrac{1}{3} lim_{x \rightarrow 2} \left( \dfrac{1}{x+1} \right)\]\[ \left( -\dfrac{1}{3} \right) \left( \dfrac{1}{3} \right) \]\[-\dfrac{1}{9}\]
which means that
\[lim_{x \rightarrow 2} \dfrac{\dfrac{1}{x+1}-\dfrac{1}{3}}{x-2}=-\dfrac{1}{9}\]
If roots are floating around it, can help to multiply by the conjugate.
Find
\[\lim_{x \rightarrow 2} \dfrac{\sqrt{x+3}-1}{x+2}\]
Answer:
Again you are blocked from using the Quotient Rule for limits because the limit of the denominator is zero if you plug in -2. So try multiplying both the numerator and denominator by the conjugate of the numerator:
\[\dfrac{\sqrt{x+3}-1}{x+2}=\dfrac{(\sqrt{x+3}-1)(\sqrt{x+3}+1)}{(x+2)(\sqrt{x+3}+1)}\]
\[\dfrac{(\sqrt{x+3})^2-1}{(x+2)(\sqrt{x+3}+1)}\]
\[\dfrac{x+3-1}{(x+2)(\sqrt{x+3}+1)}\]
\[\dfrac{x+2}{(x+2)(\sqrt{x+3}+1)}\]
\[\dfrac{1}{\sqrt{x+3}+1}\]
Now try evaluating the limit of the denominator, and you will see that
\[lim_{x \rightarrow -2} (\sqrt{x+3}+1)=\sqrt{-2+3}+1\]
\[\sqrt{1}+1\]
.
This means you can apply the Quotient Rule for limits to say that
\[lim_{x \rightarrow -2} \dfrac{1}{\sqrt{x+3}+1}=\dfrac{1}{\sqrt{1}+1}\].
Now you know that
\[lim_{x \rightarrow -2} \dfrac{\sqrt{x+3}-1}{x+2}=\dfrac{1}{2}\].
Finding the Limit of a Piecewise Function
For more examples of finding the limits of piecewise functions, see One-Sided Limits.
Using the function
\[f(x)= \left( \begin{matrix} x+3 &, x \geq 1\\ x^2-4 &, x<1\end{matrix} \right) \]
find
\[lim_{x \rightarrow 1} f(x)\]
if it exists.
Answer:
If this were any limit other than as \(x \rightarrow 1\) you could plug in function values to find the limit since both pieces of the function are polynomials. But \(x=1\) is where the definition of the function changes, so instead you need to look at the limit from the left and the limit from the right. For this function.
\[lim_{x \rightarrow 1^+} f(x)= lim_{x \rightarrow 1^+}(x+3)=4 \]
and
\[lim_{x \rightarrow 1^-} f(x)=lim_{x \rightarrow 1^-}(x^2-4)=-3\]
Since those two numbers aren't the same
\[lim_{x \rightarrow 1} f(x)\]
does not exist.
For more examples of limits of piecewise-defined functions, see One-Sided Limits
Finding Limits of Exponential Functions
When you are looking for limits of exponential functions, it depends on whether it is a standard exponential function, such as
\[f(x)=e^x\]
or a composite exponential function, like
\[g(x)=e^{\sqrt{x-1}}\]
If you are looking for standard exponential function limits, see Exponential Functions for a discussion on the behavior of exponential functions.
Remember that if you have two functions \(f(x)\) and \(g(x)\), and \(f(x)\) is continuous at \(g(c)\), then:
\[lim_{x \rightarrow c} f(g(x))=f\left( lim_{x \rightarrow c} g(x) \right)\]
.
Find
\[lim_{x \rightarrow 2} e^{\sqrt{x-1}}\].
Answer:
Think of this limit as the composition of two functions,
\(f(x)=e^x\) and \(g(x)=\sqrt{x-1}\)
Then
\[f \cdot g(x)=f(g(x))=e^{\sqrt{x-1}} \]
You already know that the exponential function is continuous everywhere, and that \(g(x)\) has a limit as \(x \rightarrow 2\). Therefore
\[lim_{x \rightarrow 2} f(g(x))=f \left(lim_{x \rightarrow 2} g(x) \right)\]
\[f\left( lim_{x \rightarrow 2} \sqrt{x-1} \right)\]
\[f(\sqrt{2-1})\]
\[f(1)=e^1=e\]
Finding the Derivative of a Function by the Limit Process
You may wonder how to find the derivative of a function using limits. That is a larger topic than will fit in this article, so for more information, see Derivative Functions and Derivatives as Rates of Change.
Finding Limits of Specific Functions - Key takeaways
- Always check to see if you can apply the Limit Law properly before using it. Be especially careful with the Quotient Rule.
- When looking for the limit of a rational function, using algebra to rewrite the function can be very helpful. Also, consider multiplying by conjugates in the case of roots in the rational function.
- If you are finding the limit of a piecewise function where the function changes definition, use one-sided limits.
- For finding the limit of exponential functions, or other composite functions, remember that if you have two functions \(f(x)\) and \(g(x)\), and \(f(x)\) is continuous at \(g(c)\), then: \[lim_{x \rightarrow c} f(g(x))=f \left(lim_{x \rightarrow c}g(x) \right)\].
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