Now that we’ve found limits graphically and looked at piecewise functions, we will now look at how to find limits at points of discontinuity using algebra. In this post we will go through three distinct examples showing different techniques for finding limits. These examples will cover most of the limits we will see in the early part of Calculus.

**Rational functions**

Here we will find

\begin{align*}

\lim_{x \to 2}\frac{x^{2}+x-6}{x-2}.

\end{align*}

As we start, we want to plug \(2\) into the function inside the limit to see what happens. In this case we get

\begin{align*}

\frac{2^{2}+2-6}{2-2}=\frac{0}{0}.

\end{align*}

Therefore, the function is undefined at the point \(2\). This also means the function is discontinuous at \(x=2\), so we can’t just plug in to the find the function value. Furthermore, since we have \(\frac{0}{0}\), we can’t say whether the limit exists or if we have a vertical asymptote without more work. In this case, our work will involve factoring the top and the bottom of the function.

We then get that

\begin{align*}

\lim_{x \to 2}\frac{x^{2}+x-6}{x-2}&=\lim_{x \to 2}\frac{(x+3)(x-2)}{x-2},

\end{align*}

We now note that \(\frac{x-2}{x-2}=1\) for all \(x \neq 2\). At \(2\) this is undefined, but since we are taking a limit, we only care about what happens as we get close to, but not equal to, 2. We now see that

\begin{align*}

\lim_{x \to 2}\frac{x^{2}+x-6}{x-2}&=\lim_{x \to 2}\frac{(x+3)(x-2)}{x-2}, \\

&=\lim_{x \to 2}x+3.

\end{align*}

Now that we have cancelled out the \((x-2)\) on top and bottom, we are left with a new simpler limit. We will now want to start over and try to find this limit. As we look at \(x+3\), we note that if we plug in \(x=2\) we get \(5\). Therefore, \(x+3\) is defined and continuous at \(x=2\), so

\begin{align*}

\lim_{x \to 2}\frac{x^{2}+x-6}{x-2}&=\lim_{x \to 2}\frac{(x+3)(x-2)}{x-2}, \\

&=\lim_{x \to 2}x+3, \\

&=5.

\end{align*}

We see here that we were able to factor the top and bottom. Since we got out \(\frac{0}{0}\), we are indeed guaranteed that we can factor an \(x-2\) (in general \(x-c\)) of both. Since \(x \neq 2\) (\(x \neq c\)), we will be able to factor out one of these terms to produce a simpler limit. In this case, we were able to find this new limit quickly, but it may be the case that still more work is required.

**Fractions**

We now want to find

\begin{align*}

\lim_{x \to 0}\frac{\frac{1}{2+x}-\frac{1}{2}}{x}.

\end{align*}

As we did in the previous case, we will start by calculating the value of the function at \(x=0\). In this case, we arrive at

\begin{align*}

\frac{\frac{1}{2+0}-\frac{1}{2}}{0}&=\frac{0}{0}.

\end{align*}

That is, the function is undefined at \(x=0\). However, since we got \(\frac{0}{0}\), we know that we have an indeterminate limit and must do more work to determine what the limit is. In the last case, we were able to factor the top and bottom so that we could cancel out an \(x-2\). Here, we would like to factor out an \(x\). However, the top consists of two fractions, so we can’t just factor. Instead we need to combine these fractions using a common denominator, so that we can then simplify and factor.

In order to get a common denominator for the fractions, we multiply each by the other’s denominator. That is, we will take

\begin{align*}

\frac{1}{2+x}-\frac{1}{2}&=\frac{2}{2}\frac{1}{2+x}-\frac{1}{2}\frac{2+x}{2+x} \\

&=\frac{2}{2(2+x)}-\frac{2+x}{2(2+x)} \\

&=\frac{2-(2+x)}{2(2+x)} \\

&=\frac{-x}{2(2+x)}.

\end{align*}

Note that we have now simplified the top from one fraction into two, but we can still simplify the problem. We find that

\begin{align*}

\frac{\left(\frac{-x}{2(2+x)}\right)}{x}&=\frac{-x}{2(2+x)}\frac{1}{x} \\

&=\frac{-x}{2(2+x)x},

\end{align*}

since dividing fractions is the same as multiplying by the reciprocal.

We next note that, since we are taking the limit as \(x\) approaches \(0\), it doesn’t matter what happens exactly at \(0\). Therefore, \(\frac{x}{x}=1\) since we have \(x \neq 0\). We then find that

\begin{align*}

\lim_{x \to 0}\frac{\frac{1}{2+x}-\frac{1}{2}}{x}&=\lim_{x \to 0}\frac{-x}{2(2+x)x} \\

&=\lim_{x \to 0}\frac{-1}{2(2+x)} \\

&=\frac{-1}{4},

\end{align*}

where the last step is completed by calculating the function at \(x=0\) and noting that we get a number out.

For this example, we saw that we had two fractions either being added or subtracted from each other. In order to simplify the problem we had to combine these fractions using a common denominator. This allowed us to turn the problem into one of the first type, where we had a rational function, so that we could factor and simplify. We were then, finally, able to calculate the limit by finding the function value of a continuous function.

**Square roots**

We now want to find

\begin{align*}

\lim_{x \to 0}\frac{\sqrt{9+x}-3}{x}.

\end{align*}

We again by finding the function value at the point \(x=0\). In this case, we get

\begin{align*}

\frac{\sqrt{9+0}-3}{0}=\frac{0}{0},

\end{align*}

which again gives us that the function is undefined and that the limit is indeterminate. The difficulty with simplifying this problem is the square roots. Therefore, if we want to continue, we have to find a way to deal with these.

We note that in general if you want to change a mathematical expression, you can, as long as you undo what you do. Therefore, we may want to square the function to get rid of the square roots, then take a square later. However, if we square the top we get

\begin{align*}

\left(\sqrt{9+x}-3\right)^{2}&=9+x+2\sqrt{9+x}+9 \\

&=18+x+2 \sqrt{9+x}.

\end{align*}

We, therefore, end up with a more complicated problem, so this doesn’t look like the technique we want to use.

While our goal remains getting rid of the square root, we will try a different tactic. Here we will multiply by the top, but we will change the \(-\) between the terms to a \(+\). In so doing, we are multiplying by what is called the conjugate of the top. We then get

\begin{align*}

\left(\sqrt{9+x}-3\right)\left(\sqrt{9+x}+3\right) &=9+x-9=x.

\end{align*}

With this technique, we do arrive at a much simpler equation. If we go back to the original problem, we will now incorporate this into the equation. Recall, however, that we need to undo anything we do in the equation, so we will divide by the same thing we multiplied. This gives us

\begin{align*}

&\left(\frac{\sqrt{9+x}-3}{x}\right)\left(\frac{\sqrt{9+x}+3}{\sqrt{9+x}+3}\right) \\

&=\frac{x}{x\left(\sqrt{9+x}+3\right)}.

\end{align*}

We can now see that, since we are letting \(x\) get close to, but not equal to, 0, we will have that \(\frac{x}{x}=1\). We now have

\begin{align*}

\lim_{x \to 0}\frac{\sqrt{9+x}-3}{x}&=\lim_{x \to 0}\frac{x}{x\left(\sqrt{9+x}+3\right)} \\

&=\lim_{x \to 0}\frac{1}{\sqrt{9+x}+3} \\

&=\frac{1}{\sqrt{9+0}+3}\\

&=\frac{1}{6}.

\end{align*}

Again, we were able to find the limit by cancelling out a term that is equal to \(1\) when \(x \neq 0\). Therefore, we want to find a limit involving square roots, we can simplify by multiplying the top and bottom of the fraction by the conjugate of the term with the square root. This then allows us to cancel terms out.

**Conclusion**

We have seen three different examples of taking limits algebraically. With a rational function, we were able to factor and cancel. If there were two fractions being added or subtracted, we combined these using a common denominator then factored and simplified. Then, when given a problem with a square root, we multiplied by the conjugate then factored and simplified. In early Calculus, these three types of limits will come up quite frequently. If you can remember these three techniques, you should be able to calculate any limits for the first portion of the class.

Thanks for reading mathematicians, I hope you enjoyed the post. If you learned something, make sure to share this post on Social Media with anyone else that may find it useful.

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