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

Summary Submit Skip Question

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

We can factor using difference of squares and cancel terms:

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

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

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