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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 $$\begin{aligned} \lim_{x\to a}\frac{(x^2-x\left | x \right |-a^2)^2-a^4\left | a \right |}{x-a}&=\lim_{x\to a}(x^2-x\left | x \right |-a^2)^2+x\cdot\frac{d}{dx}(x^2-x\left | x \right |-a^2)^2\\ &=a^4-aL \end{aligned} $$

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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 $$x<0$$

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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 $L=-8a^3.$

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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.

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