Introduction

Welcome to the FiveHive article for AP Calculus AB/BC for Unit 1.6 Determining Limits Using Algebraic Manipulation!

We have everything we need to get into the nitty-gritty with limits. Now we will use algebra to evaluate and manipulate analytical/algebraic limits that are indeterminate and reveal their true results.

As usual with FiveHive articles, we will only cover what is mentioned on the CED for Unit 1.6.

What is an indeterminate limit?

An indeterminate limit is when plugging in the value the limit is approaching into the function results in one of these forms listed: 

$\frac{0}{0}$

$\frac{\infty }{\infty }$

$\infty -\infty$

$0\cdot \infty$

$0^0$

${\infty }^0$

$1^{\infty }$

These are the most common indeterminate forms and other forms can usually be transformed into them (e.g. ${\log }_{\infty }\infty$ can be transformed into $\frac{\infty }{\infty }$). This limit can still exist. We just need to manipulate the limit into an equivalent form using algebra, and later on, calculus. This topic of an indeterminate limit will come back up again at the end of Unit 4, so be prepared! 

In this lesson the only indeterminate limit type we will go over is $\frac{0}{0}$, but there are still many different problem types. 

Example 1: 

Calculate $\underset{x\to 2}{\lim }\frac{x^2-4}{x-2}$. 

Plugging in $x=2$ into the function gives us $\frac{0}{0}$. The limit can still exist though.  The $x^2-4$ term in the numerator can be factored into $(x+2)(x-2)$. Now the limit is

$\underset{x\to 2}{\lim }\frac{(x+2)\cancel{(x-2)}}{\cancel{x-2}}$

$=\underset{x\to 2}{\lim }x+2=4$. 

$\underset{x\to 2}{\lim }\frac{x^2-4}{x-2}=4$.

In a similar vein, there is also square root conjugate multiplication.

Example 2:

Calculate $\underset{x\to -7}{\lim }\frac{x+7}{(\sqrt{x+16}-3)}$

Once again this is an indeterminate $\frac{0}{0}$ limit. Even though $x+7$ can be specially factored, I think it is easier to multiply by the square root conjugate. 

The conjugate here is $\sqrt{x+16}+3$. 

$\underset{x\to -7}{\lim }\frac{(x+7)(\sqrt{x+16}+3)}{(\sqrt{x+16}-3)(\sqrt{x+16}+3)}$. 

Notice how this is like a “reverse” difference of squares factoring. I highly recommend NOT to multiply out the numerator, you will just further confuse yourself with that algebra. Instead, multiplying the bottom gives 

$\underset{x\to -7}{\lim }\frac{(x+7)(\sqrt{x+16}+3)}{x+16-9}$ for $x\ge -16$. 

The value we are approaching satisfies that inequality ($-7\ge -16$), so we can continue. 

$\underset{x\to -7}{\lim }\frac{\cancel{(x+7})(\sqrt{x+16}+3)}{\cancel{x+7}}$

$=\underset{x\to -7}{\lim }\sqrt{x+16}+3=3+3=6$.

$\underset{x\to -7}{\lim }\frac{x+7}{(\sqrt{x+16}-3)}=6$.

Finally, dust off your trig skills. You will need a bit of algebra and trigonometry for this next limit.

Example 3:

Calculate $\underset{x\to \frac{\pi }{3}}{\lim }\frac{{\cos }^{2}x-\frac{1}{4}}{\tan x-\sqrt{3}}$.

Once again this is $\frac{0}{0}$ (hopefully you got your trigonometric exact values right). 

$\underset{x\to \frac{\pi }{3}}{\lim }\frac{{\cos }^{2}x-\frac{1}{4}}{\tan x-\sqrt{3}}$

$=\underset{x\to \frac{\pi }{3}}{\lim }\frac{({\cos }^{2}x-\frac{1}{4})(\tan x+\sqrt{3})}{{\tan }^{2}x-3}$

$=2\sqrt{3}\underset{x\to \frac{\pi }{3}}{\lim }\frac{{\cos }^{2}x-\frac{1}{4}}{{\tan }^{2}x-3}$

$=2\sqrt{3}\underset{x\to \frac{\pi }{3}}{\lim }\frac{{\cos }^{2}x-\frac{1}{4}}{\frac{{\sin }^{2}x-3{\cos }^{2}x}{{\cos }^{2}x}}$

$=2\sqrt{3}\underset{x\to \frac{\pi }{3}}{\lim }\frac{1}{4}\frac{{\cos }^{2}x-\frac{1}{4}}{{\sin }^{2}x-3{\cos }^{2}x}$

$=\frac{\sqrt{3}}{2}\underset{x\to \frac{\pi }{3}}{\lim }\frac{\frac{3}{4}{\cos }^{2}x+\frac{1}{4}{\cos }^{2}x-\frac{1}{4}}{{\sin }^{2}x-3{\cos }^{2}x}$

$=\frac{\sqrt{3}}{2}\underset{x\to \frac{\pi }{3}}{\lim }-\frac{1}{4}\frac{\cancel{{\sin }^{2}x-3{\cos }^{2}x}}{\cancel{{\sin }^{2}x-3{\cos }^{2}x}}$

$=-\frac{\sqrt{3}}{8}$

$\underset{x\to \frac{\pi }{3}}{\lim }\frac{{\cos }^{2}x-\frac{1}{4}}{\tan x-\sqrt{3}}=-\frac{\sqrt{3}}{8}$.

Practice

That was a lot of work, but you finally got through it! Now it is time for some independent practice using algebra and trigonometry to solve some indeterminate limits.