Question

How would you solve #12?

Answer 1

You can use the difference of square equation:

`a^2-b^2 = (a+b)(a-b)`

In fact,
`x^4` is the square of
`x^2` and
`a^4` is the square of
`a^2`

This means that you can write
`x^4-a^4 = (x^2+a^2)(x^2-a^2)`

So, the denominator simplifies:

`(x^4-a^4)/(x^2-a^2) = ((x^2+a^2)(x^2-a^2))/(x^2-a^2) = x^2+a^2`

So, when
`x \to a`, obviously
`x^2 \to a^2`, and the limit is simply

`\displaystyle \lim_(x \to a) (x^4-a^4)/(x^2-a^2) = \lim_(x \to a) x^2+a^2 = a^2+a^2 = 2a^2`