For which values of $\alpha\in\mathbb{R}$, does the limit

$$\lim \limits_{x \to \alpha}\frac{(x^3-2\alpha x^2+\alpha^4 x)\ln\left\lvert x\right\rvert}{(x-1)(x-\alpha)^2}$$

exist ?

Can someone explain me how to figure out the values of $\alpha$ in a general case and how to find those values in this case.

Thanks in advance.

Edit: in my case it s $\alpha^4 x $. But still doesn't really change the problem.

  • $\begingroup$ Can you use the rules of L'Hospital? $\endgroup$ – Dr. Sonnhard Graubner Mar 16 '18 at 15:09
  • $\begingroup$ I dont think I can use l'Hospital. $\endgroup$ – Sami Mir Mar 16 '18 at 15:14
  • $\begingroup$ We should simplify $(x-\alpha)^2$, so we need $\alpha$ to be a double root of $x^3-2\alpha x^2+\alpha x=x(x^2-2\alpha x+\alpha )$. Thus $\alpha=0$ or $1$. $\endgroup$ – Qurultay Mar 16 '18 at 15:14

If $\alpha=1$, the numerator is $$x (x^2-2x+1)\ln (|x|)=$$ $$x (x-1)^2\ln (|x|) $$

the limit is then

$$\lim_{x\to 1}\frac {x\ln (x)}{x-1}=$$ $$\lim_{y\to 0}\frac {(y+1)\ln (y+1)}{y}=1$$

if $\alpha \ne 1$, the numerator $\to$ $$\alpha^2 (1-\alpha) \ln(|\alpha|)$$

If $|\alpha|>1$ the limit is $-\infty$

If $|\alpha|<1$, it is $+\infty $.

it belongs to you now to see the case $\alpha=-1$.


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