Basic Limit Properties Question 4

If $a\neq 0$, then $\lim\limits_{x\to a} \dfrac{x^2-a^2}{x^4-a^4} $ is

$\dfrac{1}{a^2} $

$\dfrac{1}{2a^2} $

$0 $

nonexistent

Approach 1

Substituting in the value $x=a$ results in the indeterminate form $\dfrac{0}{0}$.

We can factor using difference of squares and cancel terms:

$$ \lim\limits_{x\to a} \frac{x^2-a^2}{x^4-a^4} $$ $$ = \lim\limits_{x\to a} \frac{x^2-a^2}{(x^2+a^2)(x^2-a^2)} $$ $$ = \lim\limits_{x\to a} \frac{1}{x^2+a^2} $$ $$ =\boxed{\frac{1}{2a^2}} $$

Approach 2

Once you know L'Hospital's rule, you can use if for all indeterminate forms:

$$ \lim\limits_{x\to a} \frac{x^2-a^2}{x^4-a^4} $$ $$ = \lim\limits_{x\to a} \frac{2x}{4x^3} $$ $$ = \frac{2a}{4a^3} $$ $$ = \boxed{\frac{1}{2a^2}} $$