Limit Calculation $\lim_{x\to \infty} x\log(x^2+x)- x^2\log(x +1)$

Question

I am having a problem with the calculation of the following limit. I need to find

$$\lim_{x\to \infty} x\log(x^2+x)- x^2\log(x +1).$$

I've been trying in this way but I'm not sure if it is correct:

$$\lim_{x\to \infty} x\log(x(x+1))- x^2\log(x +1)$$

$$\lim_{x\to \infty} x\log(x)+x\log(x+1)- x^2\log(x +1)$$ $$\lim_{x\to \infty} x\log(x)+\lim_{x\to \infty}\log(x+1) (x- x^2)$$

the first one it should be $+\infty $ , How could I calculate the second one?

Answer 1

$\lim_\limits{x\to \infty} x\log(x^2+x)- x^2\log(x +1)\\ \lim_\limits{x\to \infty} \log\frac{(x^2+x)^x}{(x +1)^{x^2}}\\ \lim_\limits{x\to \infty} \log\frac{x^x}{(x+1)^{x^2-x}}\\ $

The denominator of that fraction is growing much faster than the numerator. As $x\to\infty$ the limit approaches $\log 0 = -\infty$

Answer 2

You can try the substitution $x=1/t$ that brings the limit into $$ \lim_{t\to0^+}\left( \frac{1}{t}\log\left(\frac{1}{t^2}+\frac{1}{t}\right) -\frac{1}{t^2}\log\left(1+\frac{1}{t}\right) \right) = \lim_{t\to0^+} \frac{t\log(1+t)-2t\log t-\log(1+t)+\log t}{t^2} $$ Since $\lim_{t\to0^+}\log(1+t)=0$ and also $\lim_{t\to0}t\log t=0$, the limit is $-\infty$.

Answer 3

For large $x$, $x\log(x^2+x)$ behaves like $2x\log x$ while $x^2\log(x+1)$ behaves like $x^2\log(x)$, hence the limit is trivially $-\infty$.