# Evaluate $\lim_{x \to -\infty} \ln(-x^3+x)$

Evaluate $$\lim_{x \to -\infty} \ln(-x^3+x).$$

I was wondering if I can solve this limit in this way: $$\lim_{x \to -\infty} \ln(-x^3+x)=\lim_{x \to -\infty} \ln\left[x^3\left(1+\frac{1}{x^2}\right)\right].$$

At this point, I just considered $\ln(x^3)$ because $1$ doesn't make any difference and $1/x^2$ tends towards $0.$ So, the result will be $0.$ And I found it because I know the graph of the logarithm of $x$ to the power of an odd number. So, my second question is, is it possible to understand what the result of $\lim_{x\to - \infty} \ln(x^3)$ algebraically without thinking of the graph?

Any suggestion and help will be appreciated. Thank you in advance and have a good day :)

• what happen to - sign – King Tut Jan 20 '18 at 17:21
• this limit makes no sense! – Dr. Sonnhard Graubner Jan 20 '18 at 17:27
• If $x$ is $<0$, $x^3$ is also $<0$. – hamam_Abdallah Jan 20 '18 at 17:37
• If you are ok, you can accept the answer and set as solved. Thanks! – user Jan 22 '18 at 21:58

The correct way is

$$\lim_{x \to -\infty} \ln(-x^3+x)=\lim_{x \to -\infty} \ln\left[-x^3\left(1-\frac{1}{x^2}\right)\right]=\lim_{x \to -\infty} \left[\ln -x^3+\ln\left(1-\frac{1}{x^2}\right)\right]=\\=\lim_{x \to -\infty} \ln -x^3+\lim_{x \to -\infty} \ln\left(1-\frac{1}{x^2}\right)=+\infty+0=+\infty$$

indeed

$$\lim_{x \to -\infty} \ln -x^3=\lim_{x \to -\infty} 3\ln -x=\lim_{y \to +\infty} 3\ln y=+\infty$$

$y:=-x.$

Consider :

$\lim_{ y \rightarrow +\infty}\ln(y^3-y).$

$\ln(y^3-y) =$

$\ln(y) +\ln(y-1) + \ln(y+1).$

Hence :

$\lim_{y \rightarrow +\infty} \ln(y^3-y)$ is?

hint

For $x <-1,$

$$-x^3+x=(-x)^3(1-\frac {1}{x^2})$$

$$\ln (-x^3+x)=3\ln (-x)+\ln (1-\frac {1}{x^2})$$

$$\lim_{x \to -\infty} \ln(-x^3+x)=\lim_{x \to \infty}\ln(x^3-x)$$ $$=\lim_{x \to \infty}\ln(x(x^2-1))=\lim_{x \to \infty}\ln(x(x-1)(x+1))=$$ $$=\lim_{x \to \infty}\big(\ln(x)+\ln(x-1)+\ln(x+1)\big)$$

and all the three summands tend to $\infty$.