What is wrong with this solution to the limits question? Evaluate $\lim\limits_{x\to\infty}x-x^2\ln\bigg(1+\dfrac1{x}\bigg)$.
My solution:
\begin{align}
\lim\limits_{x\to\infty}x-x^2\ln\bigg(1+\frac1{x}\bigg)&=\lim\limits_{x\to\infty}x-\frac{x\ln(1+\frac1{x})}{\frac{1}x}\\
&=\lim\limits_{x\to\infty}x-(x\cdot 1)\\
&=\lim\limits_{x\to\infty}x-x\\
&=0
\end{align}
I got $0$ as answer, but the correct answer is $\frac12$.
I solved it using another method, but I just need to know why this won't work. Any ideas?
 A: When you went from 
$$  \lim_{x \rightarrow \infty} x - \left( x \frac{\ln \left( 1+\frac{1}{x} \right)}{\frac{1}{x}} \right)  $$
to 
$$  \lim_{x \rightarrow \infty} x - \left( x \cdot 1 \right)  $$
you must have passed through an expression where "$ \lim_{x \rightarrow \infty} \frac{\ln \left( 1+\frac{1}{x} \right)}{\frac{1}{x}}$" appeared.  (Otherwise, how did you replace that subexpression with its limit?)
Pretty much the only way to do that is along \begin{align*}
\lim_{x \rightarrow \infty} \; & \left( x - \left( x \frac{\ln \left( 1+\frac{1}{x} \right)}{\frac{1}{x}} \right) \right)  \\
\qquad &\overset{?}{=} \left( \lim_{x \rightarrow \infty} x \right) - \left( \lim_{x \rightarrow \infty} x \right) \left( \lim_{x \rightarrow \infty} \frac{\ln \left( 1+\frac{1}{x} \right)}{\frac{1}{x}} \right)  \text{,}  \end{align*}
which is hopeless, because two of those limits do not exist.  So the equality can (and as you have found, does) fail.
Always remember, the various versions of 
$$  \lim_{\dots} {\dots} = \left( \lim_{\dots} \dots \right) [\text{operation}] \left( \lim_{\dots} \dots \right) $$
all require that the limits on the right exist for equality to be guaranteed.
A: $$
\begin{align}
\lim_{x\to\infty}\left(x-x^2\log\left(1+\frac1{x}\right)\right)
&=\lim_{x\to\infty}\left(x-\frac{x\log\left(1+\frac1x\right)}{\frac1x}\right)\tag1\\
&=\lim_{x\to\infty}\left(x-x\lim_{x\to\infty}\frac{\log\left(1+\frac1x\right)}{\frac1x}\right)\tag2\\
&=\lim_{x\to\infty}x\lim_{x\to\infty}\left(1-\frac{\log\left(1+\frac1x\right)}{\frac1x}\right)\tag3\\
\end{align}
$$
Step $(1)$ is fine. Step $(2)$ is where the argument in the question goes wrong. It is not legal to simply apply a limit to a piece inside another limit. Step $(3)$ is fine and shows that step $(2)$ introduced a limit of the form $\infty\cdot0$.
A: Write your term in the form
$$\frac{\frac{1}{x}-\ln\left(1+\frac{1}{x}\right)}{\frac{1}{x^2}}$$
and use the rules of L'Hospital.
A: $$L=\lim_{x\rightarrow \infty} x\left(1-x\ln(1+1/x) \right) \Rightarrow x (1-x(\frac{1}{x}-\frac{1}{2} \frac{1}{x^2} + \frac{1}{3} \frac{1}{x^3}..))=\lim_{x \rightarrow \infty}(\frac{1}{2}-\frac{1}{3} \frac{1}{x})=\frac{1}{2}.$$
