Please identify my mistake: $\lim_{h\to0}\frac{\ln (1+2 h)-2 \ln (1+h)}{h^2}$

Question

Anyone please tell what mistake I am making?

$\begin{aligned} \lim_{h \to 0} & \frac{\ln (1+2 h)-2 \ln (1+h)}{h^2} \\ & =\frac{4 \ln (1+2 h)}{4 h^2}-\frac{2 \ln (1+h)}{h^2} \\ & =\frac{4 \ln (1+2 h)}{(2 h)(2 h)}-\frac{2\ln (1+h)}{(h)(h)} \\ & =\frac{4}{2 h}-\frac{2}{h} \\ & =0 \end{aligned}$

I want to know why I am getting zero.

Answer 1

The expression $\lim_{h \to 0} \frac{ \ln(1+2h) - 2 \ln(1+h)}{h^2}$ means the limit of the expression inside as $h$ is a real number tending to $0.$ The scope of definition of $h$ is restricted to the expression of the limit. However, the expression in the second line $\frac{4 \ln(1+2h)}{4h^2}$ has no meaning because $h$ has no declared meaning here.

Next, the addition rule of limits states that the limit of sum of two expressions is the sum of the limits of two expressions provided both exist and are well defined. In the case you are considering, neither $\lim_{h \to 0}\frac{4\ln(1+2h)}{4h^2}$ nor $\lim_{h \to 0} \frac{2 \ln(1+h)}{h^2}$ is defined as a real number.

Finally, even if the limits both exist and are equal to $+\infty,$ you appear to be computing $+\infty-\infty=0$ which is invalid because the operation of addition is defined on pairs of real numbers and infinity is not a real (or complex) number.

Answer 2

As has been said, you are taking the limit in some parts of the expression only, then simplifying to take another limit later: won't work.

However this can be fixed. Here you have replaced parts of the expression with approximations. It's correct if you take the rest into account.

That is, you are allowed to write $\log(1+2h)=2h+O(h^2)$, therefore $\dfrac{\log(1+2h)}{h^2}=\dfrac{2h+O(h^2)}{h^2}$, then likewise $\log(1+h)=h+O(h^2)$ therefore $2\dfrac{\log(1+h)}{h^2}=\dfrac{2h+O(h^2)}{h^2}$.

And then the difference is

$$\frac{\log(1+2h)}{h^2}-\frac{2\log(1+h)}{h^2}=\frac{2h+O(h^2)-2h-O(h^2)}{h^2}=\frac{O(h^2)}{h^2}=O(1)$$

That is, all you can conclude is the LHS is bounded. Note the $O(h^2)$ don't cancel, all you can say is $O(h^2)-O(h^2)=O(h^2)$. Think of $O(h^2)$ as "an unknown function $f$ such that $f(h)=g(h)h^2$, with bounded $g$", and each time you are using such an approximation, you have to assume it's another such unknown function.

To find the limit, take the approximation an order further: $\frac{log(1+2h)}{h^2}=2h-2h^2+O(h^3)$ and $2\log(1+h)=2h-h^2+O(h^3)$, then

$$\frac{\log(1+2h)}{h^2}-2\frac{\log(1+h)}{h^2}=\frac{2h-2h^2+O(h^3)-2h+h^2+O(h^3)}{h^2}=\frac{-h^2+O(h^3)}{h^2}=-1+O(h)$$

And the limit is $-1$.