Example 1: Complex Fraction Limit

 
$\underset{x\to 0}{\lim }\frac{7\mathrm{x}}{\frac{1}{4–\mathrm{x}}–\frac{1}{4+\mathrm{x}}}$ 

Step 1: Try Option 1: Plug it in

Always try plugging in the $x–\mathit{value}$ you are heading towards.

In this example we get division by zero, which means Option 1 has failed.

$\underset{x\to 0}{\lim }\frac{7x}{\frac{1}{4–x}–\frac{1}{4+x}}=\frac{7\left(\mathrm{(}0\mathrm{)}\right)}{\frac{1}{4–\left(\mathrm{(}0\mathrm{)}\right)}–\frac{1}{4+\left(\mathrm{(}0\mathrm{)}\right)}}=\frac{0}{0}$

Step 2: Add or Subtract Fractions

Wherever the fractions are being added or subtracted, combine them into a single fraction.

Remember the first thing to do when adding or subtracting fractions is to find a common denominator. The easiest way to do this will be to take the denominator from one fraction and multiply it on top and bottom of the other fraction, and then do it with the other fraction.

Bring the fractions together in a single fraction over the common denominator and simplify only the numerator.

Remember when subtracting equations, it is important to keep a parenthesis around the entire piece you are subtracting, and then distribute the negative sign through the entire subtracted piece.

Then combine like terms.

$\underset{x\to 0}{\lim }\frac{7x}{\frac{(4+x)}{(4+x)} \frac{1}{4–x}–\frac{1}{4+x}\frac{(4–x)}{(4–x)}}=$

$\underset{x\to 0}{\lim }\frac{7x}{ \frac{(4+x)}{(4+x)(4–x)}–\frac{1(4–x)}{(4+x)(4–x)}}=$

$\underset{x\to 0}{\lim }\frac{7x}{ \frac{(4+x)–1(4–x)}{(4+x)(4–x)}}=$

$\underset{x\to 0}{\lim }\frac{7x}{ \frac{4+x–4+x}{(4+x)(4–x)}}=$

$\underset{x\to 0}{\lim }\frac{7x}{ \frac{2x}{(4+x)(4–x)}}=$

Step 3: Apply the Division of Fractions Rewrite.

Different people have different sayings, but one of the most common division of fractions rewrite sayings goes, “Copy Dot Flip”. Copy the top fraction, put a Dot for multiplication, then Flip the bottom fraction.

Remember that everything is a fraction, as you can always put any value over 1.

So, $7x=\frac{7\mathrm{x}}{1}$.

COPY   Dot    Flip

$\underset{x\to 0}{\lim }\frac{7\mathrm{x}}{1}\mathbf{\cdot }\frac{\left(\mathrm{(}4–\mathrm{x}\mathrm{)}\right)\left(\mathrm{(}4+\mathrm{x}\mathrm{)}\right)}{2\mathrm{x}}=$

Step 4: Simplify the equation.

You will usually find that you will get some convenient cancelation between the top and bottom of your fractions, and that that simplification is just what you needed to finish up the problem. (Finally)

In this example we are able to cancel an $x$ on the top and bottom of the fraction.

$\underset{x\to 0}{\lim }\frac{7\overline{)x}}{1}\mathbf{\cdot }\frac{\left(\mathrm{(}4–x\mathrm{)}\right)\left(\mathrm{(}4+x\mathrm{)}\right)}{2\overline{)x}}=$

$\underset{x\to 0}{\lim }\frac{7}{1}\mathbf{\cdot }\frac{\left(\mathrm{(}4–x\mathrm{)}\right)\left(\mathrm{(}4+x\mathrm{)}\right)}{2}=$

Step 5: Back to Option 1: Plug it in.

$\underset{x\to 0}{\lim }\frac{7\left(\mathrm{(}4–\mathrm{x}\mathrm{)}\right)\left(\mathrm{(}4+\mathrm{x}\mathrm{)}\right)}{2}=\frac{7\left(\mathrm{(}4–\left(0\mathrm{)}\right)\right)\left(\mathrm{(}4+\left(0\mathrm{)}\right)\right)}{2}=\frac{7\cdot 16}{2}=56$

Final Result:

$\underset{x\to 0}{\lim }\frac{7\mathrm{x}}{\frac{1}{4–\mathrm{x}}–\frac{1}{4+\mathrm{x}}}=56$

Meaning:

The overall limit as $x$  approaches $0$ of $\frac{7\mathrm{x}}{\frac{1}{4–\mathrm{x}}–\frac{1}{4+\mathrm{x}}}$ is $y=56$.