Find the required limit algebrically, if it exists. lim (x+3)/(x-3) if x approaches 4 ?

Jun 18, 2018

The limit exists and is equal to 7

Explanation:

I am not sure exactly what is meant by "algebraically" in the question. So I will try to show this by a couple of different approaches.

Method 1

The simplest approach would be to use the properties of continuous functions. A standard result is that the ratio f(x)(/g(x) of two continuous functions $f \left(x\right)$ and $g \left(x\right)$ is continuous at $x = a$, provided $g \left(a\right) \ne 0$.

Since $x + 3$ and $x - 3$ are obviously continuous functions (you may prove this statement, if proof is needed, using the property that the sum of continuous functions is continuous), and $x - 3$ does not vanish at $x = 4$, the ratio $\frac{x + 3}{x - 3}$ is continuous at $x = 4$.

Hence, the limit as $x \to 4$ of $\frac{x + 3}{x - 3}$ is equal to the value of $\frac{x + 3}{x - 3}$ at $x = 4$, namely $\textcolor{red}{\frac{4 + 3}{4 - 3} = 7}$

Method 2

Another approach could be to use the $\epsilon - \delta$ definition of the limit. To show that the limit is 7 according to this definition we have to show that

$\forall \epsilon > 0 , \exists \delta > 0 : | x - 4 | < \delta \implies | \frac{x + 3}{x - 3} - 7 | < \epsilon$

Now

$| \frac{x + 3}{x - 3} - 7 | = | \frac{- 6 x + 24}{x - 3} | = 6 | x - 4 \frac{|}{|} x - 3 |$

For $| x - 4 | < \delta < \frac{1}{2}$ we have $| x - 3 | > \frac{1}{2}$ and so

$| \frac{x + 3}{x - 3} - 7 | < 6 \frac{\delta}{\frac{1}{2}} = 12 \delta$

So, for any $\epsilon > 0$ we will have

$| \frac{x + 3}{x - 3} - 7 | < \epsilon$

if we choose

$| x - 4 | < \delta = \min \left\{\frac{\epsilon}{12} , \frac{1}{2}\right\}$