How do you evaluate this limit? $\ln({x^3+2x^2+x})+ \frac{2}{x}$ How do you evaluate the following?
$$\lim_{x \to 0^+} \left [ \ln({x^3+2x^2+x})+ \frac{2}{x} \right ]$$
If I plug in $x$, I get $\infty-\infty$, which is undetermined, and I haven't been able to get the limit at a more manageable form. Can you please help me?
 A: $\ln(x^3 + 2x^2 + x) + \dfrac{2}{x} = -ln\left(\dfrac{1}{x}\right) + 2\ln(x+1) + \dfrac{2}{x} = \dfrac{1}{x} + 2\ln(x+1) + y\left(1 - \dfrac{lny}{y}\right) \to +\infty$ as $y = \dfrac{1}{x} \to +\infty$ when $x \to 0^+$, and $\dfrac{lny}{y} \to 0$.
A: Lets put $x = 1/y$ so that $y \to \infty$ as $x \to 0^{+}$. Then we can see that
\begin{align}
f(x) &= \log(x^{3} + 2x^{2} + x) + \frac{2}{x}\notag\\
&= \log\left(e^{2y}\left(\frac{1}{y^{3}} + \frac{2}{y^{2}} + \frac{1}{y}\right)\right)\notag\\
&> \log\left(e^{2y}\left(\frac{1}{y^{3}} + \frac{2}{y^{3}} + \frac{1}{y^{3}}\right)\right)\notag\\
&= \log\left(\frac{4e^{2y}}{y^{3}}\right)\tag{1}
\end{align}
Next we can see that $e^{t} \geq 1 + t > t$ for all $t$. And therefore $(e^{t})^{4} > t^{4}$. Putting $t = y/2$ we can see that $$e^{2y} > \frac{y^{4}}{16}\tag{2}$$ It follows from $(1)$ and $(2)$ that $$f(x) > \log\left(\frac{y}{4}\right)$$ Since $y \to \infty$ it follows that $\log(y/4) \to \infty$ and hence $f(x) \to \infty$ as $x \to 0^{+}$.
