Question

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

Answer 1

`10`

Explanation:

`\lim_{x\to 5}\frac{x^2-25}{x-5}`

`=\lim_{x\to 5}\frac{(x+5)(x-5)}{x-5}`

`=\lim_{x\to 5}(x+5)`

`=5+5`

`=10`

Answer 2

`10`

Explanation:

`lim_(xto5)(x^2-25)/(x-5)`

`=lim_(xto5)(cancel((x-5))(x+5))/cancel((x-5))`

`=lim_(xto5)x+5=5+5=10`

Answer 3

`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 `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

`((x+5)(x-5))/(x-5)`

Common factors cancel, and we're left with

`lim_(xto5) x+5=5+5=color(blue)(10)`

Calculus Way:

We can use L'Hôpital's rule, which says if we want to evaluate the limit of `f(x)/g(x)`, this equal to

`lim_(xtoc) (f'(x))/(g'(x))`

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

In our example,

`f(x)=x^2-25=>f'(x)=2x`

`g(x)=x-5=>g'(x)=1`

We now have the expression

`lim_(xto5)=(2x)/1=2*5=color(blue)(10)`

Both ways, we get `10`.

Hope this helps!