How would you solve #12?
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You can use the difference of square equation:
[tex] a^2-b^2 = (a+b)(a-b) [/tex]
In fact, [tex] x^4 [/tex] is the square of [tex] x^2 [/tex] and [tex] a^4 [/tex] is the square of [tex] a^2 [/tex]
This means that you can write [tex] x^4-a^4 = (x^2+a^2)(x^2-a^2) [/tex]
So, the denominator simplifies:
[tex] \dfrac{x^4-a^4}{x^2-a^2} = \dfrac{(x^2+a^2)(x^2-a^2)}{x^2-a^2} = x^2+a^2 [/tex]
So, when [tex] x \to a [/tex], obviously [tex] x^2 \to a^2 [/tex], and the limit is simply
[tex] \displaystyle \lim_{x \to a} \dfrac{x^4-a^4}{x^2-a^2} = \lim_{x \to a} x^2+a^2 = a^2+a^2 = 2a^2 [/tex]