Which limit is correct? Why does L'Hospital's rule yield a different result? https://i.stack.imgur.com/cIf8R.jpg
I would appreciate it if someone could explain why the two methods yield different results to the same limit. Have I done something incorrect? I can't seem to find the error in either method. And it would be much appreciated if the correct answer was pointed out as well.
 A: On the left side, you cannot replace $\ln(1+x)/x^2$ by $1/x$, because it is not a factor in your expression, but an addend. You would have to make sure that you are not missing terms that could contribute to the leading term. In your case, you would have to expand to the next term: $\ln(1+x)/x^2=1/x-1/2+\dots$. The $1/x$ cancels but the constant $-1/2$ contributes to the limit, since it is of the same order as $\ln(1+x)/x=1+\dots$. The final result is $1/2$, as on the right side. Hope this helps.
A: To simplify without error, you have to expand the numerator at order $2$ atleast, since the denomunatot is $x^2$, and all terms have to be expanded at the same order. What you did is that you replaced  $\frac{\ln(1+x)}{x^2}$ with its equivalent $\frac x{x^2}$, thereby forgetting that equivalence of function is not compatible with addition or subtraction.
Here is how I would have done it:

*

*$(1+x)\ln(1+x)=(1+x)\Bigl(x-\frac{x^2}2+o(x^2)\Bigr)\\=x+x^2+o(x^2)-\smash{\underbrace{\frac{x^2}2-\frac{x^3}2++xo(x^2)}_{o(x^2)}}\\[1ex]=x+\frac{x^2}2+o(x^2) $

*Therefore, the fraction is
$$\frac{x+\frac{x^2}2+o(x^2)-x}{x^2}=\frac{\frac{x^2}2+o(x^2)}{x^2}=\frac12+o(1).$$
A: $\require{cancel}$
\begin{align}
& \lim_{x\to0} \left( \frac{\log(1+x)}{x^2} - \frac 1 x \right) \\[10pt]
= {} & \lim_{x\to0} \left( \left(\frac{\log(1+x)} x \right) \cdot \frac 1 x - \frac 1 x \right) \\[12pt]
= {} & \xcancel {\left( \lim_{x\to0} \frac{\log(1+x)} x \right)\bullet \left( \lim_{x\to0} \frac 1 x \right) - \left( \lim_{x\to0} \frac 1 x \right)} \quad \text{?} \\
& \text{This does not work because those} \\
& \text{last two limits are not finite.} \\[15pt]
= {} & \xcancel {\left( \lim_{x\to0} \frac{\log(1+x)} x \right)\bullet \left( \lim_{x\to0} \left( \frac 1 x - \frac 1 x \right) \right)} \quad \text{?} \\
& \text{This does not work because } \frac 1 x - \frac 1 x \text{ is not} \\
& \text{what that first fraction what multiplied by.}
\end{align}
