Evaluating the limit of a function I just solved the following limit, to which the answer is $0$ to me, but I couldn't help but make use of L'Hôpital's Rule.
$$\lim_{x \to \infty} \left(x^2 - x \log(1+\mathrm{e}^x)\right)$$
Can someone suggest an approach that doesn't involve L'Hôpital's Rule?
 A: This is clearly equivalent to showing that $x(x - \log(1 + e^x)) \to 0$, or equivalently that $\log y\lvert \log y - \log(1 + y)\rvert \to 0$ (you can think of $x$ as $\log y$, for instance). 
A common definition of $\log y$ is $\displaystyle\int_1^y \frac{1}{x} \mathrm{d}x$. Then
$$\lim_{y \to \infty} \lvert \log y - \log(1 + y)\rvert = \left\lvert \lim_{y \to \infty} \int_1^y \frac{1}{x} - \int_1^{y+1} \frac{1}{x} \mathrm dx\right\rvert = \lim_{y \to \infty} \int_y^{y+1} \frac{1}{x}\mathrm d x < \lim_{y \to \infty} 1 \cdot \frac{1}{y},$$
so that
$$\lim_{y \to \infty} \log y \lvert \log y - \log(1 + y)\rvert < \lim_{y \to \infty} \frac{\log y}{y} \to 0,$$
which we wanted to show. $\diamondsuit$
A: We have
$$\log(1+e^x) = x + \log(1+e^{-x}) \in [x,x+e^{-x}]$$
hence
$$ x^2-x\log(1+e^x) \in [-xe^{-x},0] $$
so
$$\lim_{x\to +\infty}\left(x^2-x\log(1+e^x)\right) = 0$$
by squeezing.
A: $\log(1+e^x) = \log[e^x(1+e^{-x})] = x + \log(1+e^{-x})$
You can then use the power series for Log (substitute in $e^{-x}$) to get that
$x\log(1+e^x) \approx x(x+e^{-x})$ for large $x$.
Then the resulting limit is just:
$$\lim_{x\rightarrow \infty} xe^{-x} $$ which indeed is zero as $e^{-x}$ decays much faster than $x$
A: Using power series...
Instead of investigating the limit of the original function, which we call $g(x)$, we can investigate the limit (as $x\rightarrow\infty$) of $\exp(g)$. From this it isn't difficult to see that the result will follow if we show that $\exp(x^2)/\exp(\exp(x))\rightarrow 0$, or that $\exp/\exp(\exp(x)/x^2)\rightarrow 0$ as $x\rightarrow\infty$. Thus it is enough to show that $\exp(x)/x^2\rightarrow\infty$ as $x\rightarrow\infty$. This easily follows from the power series expansion of $\exp(x)$. 
By the way, power series is an overkill here, but it works. Notice that asking whether $\exp(x)/x$ goes to infinity, is essentially the same as asking whether $\log(x)/x$ goes to zero (see the answer by mixedmath below, which comes down to having to show just that).
A: You could sandwich the function:
$-\frac{1}{x}<x^2-x \ln(1+e^x) < x^2-x\ln(e^x) =0$
Since both left and right sides converge to $0$ so must the middle part.
A: You should be able to see that when $x$ is large then the function $\ln(1+e^x)\sim_{x\to \infty } \ln(e^x)=x $ which implies 

$$ x^2 - x\ln( 1 + e^x ) \sim_{x\to \infty} x^2 - x.x =0. $$

Added: Note that by Taylor series we have
$$ \ln(1+e^{x}) = \ln(e^x(1+e^{-x})) = x+\ln(1+e^{-x}) = x+(e^{-x}+O(e^{-2x})) $$ 
which implies

$$ x^2 - x\ln(1+e^x) = x^2-x(x+e^{-x}+O(e^{-2x})) = e^{-x}+O(e^{-2x})\to_{x\to \infty} 0. $$

A: Note that for large $x$ we have $e^{x} \approx 1 + e^{x}$ and hence $\log(1 + e^{x}) \approx x$ so that $x^{2} - x\log(1 + e^{x}) \approx 0$. What we need to do is to make this argument rigorous. We have $$\begin{aligned}x^{2} - x\log(1 + e^{x}) &= x\{x - \log(1 + e^{x})\}\\
&= x\{\log e^{x} - \log (1 + e^{x})\}\\
&= x(e^{x} - 1 - e^{x})\frac{1}{t}\text{ (by MVT where }e^{x} < t < 1 + e^{x})\\
&= -\frac{x}{t}\end{aligned}$$ Since $x/e^{x} \to 0$ as $x \to \infty$ it follows that $x/t$ also tends to $0$. Hence the desired limit is $0$.
The fundamental limit $x/e^{x} \to 0$ is easily derived if we use the series representation of $e^{x}$. Clearly $e^{x} > 1 + x + (x^{2}/2)$ and hence $0 < x/e^{x} < \dfrac{x}{1 + x + \dfrac{x^{2}}{2}}$ and using squeeze theorem we know that $x/e^{x} \to 0$ as $x \to \infty$.
