# What is lim_(x->5) (x^2-25)/(x-5) ?

$10$

#### Explanation:

$\setminus {\lim}_{x \setminus \to 5} \setminus \frac{{x}^{2} - 25}{x - 5}$

$= \setminus {\lim}_{x \setminus \to 5} \setminus \frac{\left(x + 5\right) \left(x - 5\right)}{x - 5}$

$= \setminus {\lim}_{x \setminus \to 5} \left(x + 5\right)$

$= 5 + 5$

$= 10$

Jul 18, 2018

$10$

#### Explanation:

${\lim}_{x \to 5} \frac{{x}^{2} - 25}{x - 5}$

$= {\lim}_{x \to 5} \frac{\cancel{\left(x - 5\right)} \left(x + 5\right)}{\cancel{\left(x - 5\right)}}$

$= {\lim}_{x \to 5} x + 5 = 5 + 5 = 10$

Jul 18, 2018

$10$

#### Explanation:

There are a few ways we can tackle this- I will show an algebraic way and a more calculus-themed way:

Algebraic Way:

If we plug $5$ into our expression, we immediately see that we will get indeterminate form, or $\frac{0}{0}$.

The key is to realize that we have a difference of squares in the numerator. This allows us to rewrite the expression as

$\frac{\left(x + 5\right) \left(x - 5\right)}{x - 5}$

Common factors cancel, and we're left with

${\lim}_{x \to 5} x + 5 = 5 + 5 = \textcolor{b l u e}{10}$

Calculus Way:

We can use L'HÃ´pital's rule, which says if we want to evaluate the limit of $f \frac{x}{g} \left(x\right)$, this equal to

${\lim}_{x \to c} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

We take the derivative of our numerator and denominator expression, and evaluate the limit.

In our example,

$f \left(x\right) = {x}^{2} - 25 \implies f ' \left(x\right) = 2 x$

$g \left(x\right) = x - 5 \implies g ' \left(x\right) = 1$

We now have the expression

${\lim}_{x \to 5} = \frac{2 x}{1} = 2 \cdot 5 = \textcolor{b l u e}{10}$

Both ways, we get $10$.

Hope this helps!