# Thread: Verifying a limit exists

1. ## Verifying a limit exists

Verify the following limit exists:

$\displaystyle \lim_{x \rightarrow 0} \frac{(1+\frac{x}{2})log(1+x)-x}{x^3}$
It's easy to find the limit by L'Hopital's rule, but that's the second part of the question after proving the limit exists.

Therefore, is there a way to prove that the limit exists without using L'hopitals rule (or is it acceptable to just find the limit by L'Hopitals rule and say that it must exist since we found it!)?

2. If the limit can be computed by L'Hopital then it exists.

3. Hello,
Originally Posted by Showcase_22
It's easy to find the limit by L'Hopital's rule, but that's the second part of the question after proving the limit exists.

Therefore, is there a way to prove that the limit exists without using L'hopitals rule (or is it acceptable to just find the limit by L'Hopitals rule and say that it must exist since we found it!)?
Note that $\displaystyle \ln(1+x)=x-\frac{x^2}{2}+o(x^3)$ (or big oh, I never know) when x is near 0.

Hence your limit is equivalent to :
$\displaystyle \lim_{x \to 0} \frac{\left(1+\tfrac x2\right) \left(x-\tfrac{x^2}{2}\right)-x}{x^3}$

now simplify by x :
$\displaystyle \lim_{x \to 0} \frac{\left(1+\tfrac x2\right)\left(1-\tfrac x2\right)-1}{x^2}$

now you can see that the numerator simplifies to $\displaystyle -\frac{x^2}{4}$

and then you have the limit.

as to prove why it exists, I'm not sure, maybe finding the limit is enough??

Maybe you can just say that $\displaystyle \ln(1+x) < x$ for small x. Thus $\displaystyle \ln(1+x)<x-x^2/2$ It's crafty but maybe it works ?

there surely are some mistakes in the wording... It's not my favourite stuff, so you'll do it yourself

4. Originally Posted by Moo
Hello,

Note that $\displaystyle \ln(1+x)=x-\frac{x^2}{2}+o(x^3)$ (or big oh, I never know) when x is near 0.

Hence your limit is equivalent to :
$\displaystyle \lim_{x \to 0} \frac{\left(1+\tfrac x2\right) \left(x-\tfrac{x^2}{2}\right)-x}{x^3}$

now simplify by x :
$\displaystyle \lim_{x \to 0} \frac{\left(1+\tfrac x2\right)\left(1-\tfrac x2\right)-1}{x^2}$

now you can see that the numerator simplifies to $\displaystyle -\frac{x^2}{4}$

and then you have the limit.
Moo, this is a pretty example showing why "the small o" is important. You didn't write the small o, and you obtained $\displaystyle -\frac{1}{4}$ as the limit (btw, if $\displaystyle f(x)\sim_{x\to 0^+} \ell$, this proves that $\displaystyle f(x)\to_{x\to 0^+}\ell$).

Let me do the same computation as you did, but with the o(): we have $\displaystyle \ln(1+x)=x-\frac{x^2}{2}+o(x^2)$ (or $\displaystyle \ln(1+x)=x-\frac{x^2}{2}+O(x^3)$, the conclusion is the same), so that
$\displaystyle (1+\frac{x}{2})\ln(1+x)-x=(1+\frac{x}{2})(x-\frac{x^2}{2}+o(x^2))-x$ $\displaystyle =x\left((1+\frac{x}{2})(1-\frac{x}{2}+o(x))-1\right)=x\left(-\frac{x^2}{4}+o(x)\right)$
if we divide by $\displaystyle x^3$ we get $\displaystyle -\frac{1}{4}+o\left(\frac{1}{x}\right)$. This is the sum of a constant and a function which is negligible compared to $\displaystyle \frac{1}{x}$. Since $\displaystyle \frac{1}{x}\to+\infty$, we can't conclude anything. In fact, when I wrote $\displaystyle -\frac{x^2}{4}+o(x)$, it was equivalent to write $\displaystyle o(x)$ since the first term is negligible compared to $\displaystyle x$.

The conclusion is that the expansion at the beginning is not precise enough (the neglected terms become important in the final expression). It would have been necessary to write $\displaystyle \ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3)$. Then the numerator becomes:
$\displaystyle \frac{x^4}{6}+\frac{x^3}{12}+o(x^3)=\frac{x^3}{12} +o(x^3)$,
and dividing by $\displaystyle x^3$ we get $\displaystyle \frac{1}{12}+o(1)$: the limit is $\displaystyle \frac{1}{12}$.

---

"$\displaystyle o(x^k)$" is a function which is negligible compared to $\displaystyle x^k$, so that in the expansion $\displaystyle f(x)=a_0+a_1x+\cdots+a_k x^k+o(x^k)$ each term is negligible compared to the previous ones. This shows that the constants are the "right" ones: $\displaystyle f(x)\sim a_0$, $\displaystyle f(x)-a_0\sim a_1 x$,... $\displaystyle f(x)-(a_0+a_1x+\cdots+a_{k-1}x^{k-1})\sim a_k x^k$ (provided the $\displaystyle a_i$ aren't 0).