0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $\alpha=-1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.