# Evaluate the Limit Without L'Hopital Rule

Let $$\lim_{x\to a}\frac{a^4-(x^2-x\left | x \right |-a^2)^2}{x-a}=L$$, find the value of $$\lim_{x\to a}\frac{x(x^2-x\left | x \right |-a^2)^2-a^4\left | a \right |}{x-a}$$ for $$a\neq0$$.

Using L'Hopital rule I found that the answer is $$a^4-aL$$. My question is how to solve this problem without using L'Hopital rule.

Here's my attempt using L'Hopital. \begin{aligned} \lim_{x\to a}\frac{a^4-(x^2-x\left | x \right |-a^2)^2}{x-a}&=L\\ \lim_{x\to a}\frac{d}{dx}(x^2-x\left | x \right |-a^2)^2&=-L...(1)\\ \end{aligned} Let the numerator equal to zero. $$(x^2-x\left | x \right |-a^2)^2=a^4...(2)$$ Ergo \begin{aligned} \lim_{x\to a}\frac{(x^2-x\left | x \right |-a^2)^2-a^4\left | a \right |}{x-a}&=\lim_{x\to a}(x^2-x\left | x \right |-a^2)^2+x\cdot\frac{d}{dx}(x^2-x\left | x \right |-a^2)^2\\ &=a^4-aL \end{aligned}

Hint: The numerator can be written as $$-x \left( \left| x \right| -x \right) \left( x \left| x \right| +2\, {a}^{2}-{x}^{2} \right)$$ and now distinguish the cases $$x\geq 0$$ or $$x<0$$

Case 1: $$a>0$$. Since $$x \to a$$, we can assume that $$x>0.$$ Then show that $$\frac{a^4-(x^2-x\left | x \right |-a^2)^2}{x-a}=0.$$ Hence $$L=0.$$

Case 2: $$a<0$$. Since $$x \to a$$, we can assume that $$x<0.$$ Then show that $$\frac{a^4-(x^2-x\left | x \right |-a^2)^2}{x-a}=-4x^2(x+a).$$ Hence $$L=-8a^3.$$

Write $$f(x) = (x^2-x|x|-a^2)^2$$ and note that $$f(a) = a^4$$. Now, we have that $$L = \lim_{x\to a}\frac{a^4-(x^2-x|x|-a^2)^2}{x-a} = -\lim_{x\to a}\frac{f(x)-f(a)}{x-a}= -f'(a).$$

The last equality is not due to L'Hospital rule, but the definition of derivative (which is not even important here).

The limit that you want is then

\begin{align}\lim_{x\to a}\frac{x(x^2-x|x|-a^2)^2-a^4|a|}{x-a} &= \lim_{x\to a}\frac{xf(x)-|a|f(a)}{x-a}\\ &=\lim_{x\to a}\frac{xf(x)-af(x)+af(x)-|a|f(x)+|a|f(x)-|a|f(a)}{x-a}\\ &= f(a)-|a|L+\lim_{x\to a}\frac{(a-|a|)f(x)}{x-a}.\end{align}

Observe the limit $$\lim_{x\to a}\frac{(a-|a|)f(x)}{x-a}$$. For it to be finite, $$(a-|a|)f(a) = (a-|a|)a^4$$ must be $$0$$, i.e. $$a = |a|$$.

We conclude that for $$a>0$$, the limit is $$a^4$$ (because $$L = 0$$) and for $$a<0$$ the limit doesn't exist.