So the question is,
$$\lim_{x\to 0}\{\frac{(1+x){\log(1+x)} – x}{x^2}\}$$
Method 1:
Applying l'hopital's rule the limit tend to $\bigl(\frac{1}{2}\bigr)$.
Method 2:
Solving the following way:
$$\lim_{x\to 0} \biggl(\frac{(1+x)\log(1+x)}{x^2}\biggr)-\lim_{x\to 0} \biggl(\frac{1}{x}\biggr)$$
$$\lim_{x\to 0} \biggl(\frac{1+x}{x}\biggr)\biggl(\frac{\log(1+x)}{x}\biggr)-\lim_{x\to 0} \biggl(\frac{1}{x}\biggr)$$
$$\lim_{x\to 0} \biggl(\frac{1}{x} + 1\biggr)\biggl(\frac{\log(1+x)}{x}\biggr)-\lim_{x\to 0} \biggl(\frac{1}{x}\biggr)$$
using property:
$$\lim_{x\to 0} \Biggl(\frac{\log(1+x)}{x}\Biggr) = 1$$
$$\lim_{x\to 0} \biggl(\frac{1}{x} + 1\biggr)-\lim_{x\to 0} \biggl(\frac{1}{x}\biggr)$$
$$\lim_{x\to 0} \biggl(\frac{1}{x} + 1-\frac{1}{x}\biggr)$$
$$ = 1$$
So, why is the limit different for different methods ?
Best Answer
Simply put, you cannot expand out the limit as you have done and then recombine, when the parts tend to infinity.
To see what I mean, consider the following reversed example...
$$\begin{align} 1 = \lim_{x\to0} \frac{x}x &= \lim_{x\to0} \frac{x+1}x - \lim_{x\to0} \frac1x\\ &\neq\lim_{x\to0}\frac{x+1}x(1-x) - \lim_{x\to0} \frac1x\\ &= \lim_{x\to0} \frac{1-x^2}x - \lim_{x\to0} \frac1x\\ &= \lim_{x\to0} \frac{1-x^2 - 1}x\\ &= \lim_{x\to0} \frac{-x^2}x = 0 \end{align}$$
By looking at the step with the $\neq$, we can see what is going on, here... $$ \lim_{x\to 0} \frac{x+1}x(1-x) = \lim_{x\to 0} \frac{x+1}x - \lim_{x\to 0}\ (x+1) $$ And so, multiplying $\frac{x+1}x$ by $1-x$, which has a limit of 1, has the effect of subtracting $\lim_{x\to 0} (x+1) = 1$ from the limit, thus changing the result.
The same applies here, albeit in a less obvious form. In particular, notice that $$ \frac{\log(1+x)}x \approx 1\color{red}{-\frac{x}2}+O(x^2) $$ If you examine what happens if you add this extra term (in red) to your expression, you end up with the right result.
On a more general note, splitting limits only works if the limits of the parts exist. This applies to both addition and multiplication splitting. For example, $$ \lim_{x\to0} \left(\frac1x-\frac1x\right) \neq \lim_{x\to 0} \frac1x - \lim_{x\to 0} \frac1x $$ and $$ \lim_{x\to 0} \left(\frac1xx\right) \neq \lim_{x\to 0} \frac1x \times \lim_{x\to 0} x $$