Add terms on a limit whose value depends of $\alpha$.

This may be a silly question, but I had to ask it. Yesterday, I was helping an engineer on doing a limit (we're both studying at the university and he's doing calculus now).

I helped him on doing this limit:

Solve the following limit for every $\alpha>0$: $$\displaystyle\lim_{x\to0}\frac{\ln(1+x^2)}{\rvert x\lvert^\alpha}.$$

I said him that my solution would be:

We know that in a neighborhood of $0,$ $\ln(1+x^2)\sim x^2$. So we can write: $$\displaystyle\lim_{x\to0}\frac{\ln(1+x^2)}{\rvert x\lvert^\alpha}=\lim_{x\to0}\frac{x^2}{\rvert x\lvert^\alpha}=\lim_{x\to0}\frac{\lvert x\rvert^2}{\rvert x\lvert^\alpha}\implies \begin{cases} \alpha>2 \to \displaystyle\lim_{x\to0}\frac{\ln(1+x^2)}{\rvert x\lvert^\alpha}=\infty \\ \\ \alpha<2 \to \displaystyle\lim_{x\to0}\frac{\ln(1+x^2)}{\rvert x\lvert^\alpha}=0\\ \\ \alpha=2 \to \displaystyle\lim_{x\to0}\frac{\ln(1+x^2)}{\rvert x\lvert^\alpha}=1 \end{cases}$$

But then, his professor gave them another answer, and it looked like that (I'll only post the beginning):

$$\displaystyle\lim_{x\to0}\frac{\ln(1+x^2)}{\rvert x\lvert^\alpha}=\lim_{x\to0}\frac{\ln(1+x^2)}{\rvert x\lvert^\alpha}\cdot\frac{x^2}{x^2}=\lim_{x\to0}\frac{x^2}{\lvert x\rvert^\alpha}\cdot{\ln(1+x^2)^{1/x^2}}=\\ =\lim_{x\to0}\frac{x^2}{\lvert x\rvert^\alpha}\cdot \lim_{x \to 0}{\ln(1+x^2)^{1/x^2}}=\lim_{x\to0}\frac{x^2}{\lvert x\rvert^\alpha}\text{, because (1+x^2)^{(1/x^2)}\to e.}$$

Doing maths on university, on calculus I, we've been warned several times about put on my limit this kind of fractions, even they where $x^2/x^2$ and you knew that you were multyplying by $1$ the limit. Adding terms this way inside the limits that depended on some values $\alpha$ it could lead to a problem.

So now I wanted to ask: when can I add terms on a limits on the way that his professor did? When it's not "illegal" to do it?

Thank you.

• adding where?...we're multiplying aren't we? – RE60K Mar 24 '15 at 17:36
• Sorry, in spanish it's used add as a synonym of "put inside". I'll correct it right now. – Relure Mar 24 '15 at 17:37

No it wont unless you break apart the fraction into indeterminate terms like: $$\lim_{x\to0}(x+3)=\lim_{x\to0}(x+3).\frac xx=\lim_{x\to0}(x^2+3x)\frac1x\ne\lim_{x\to0}(x^2+3x)\lim_{x\to0}\frac1x$$ It won't create a problem unless limit of $\frac{x^2}{|x|^\alpha}$ exists
• Done. From there, he found the same values for $\alpha$ that appear on my solution. – Relure Mar 24 '15 at 18:06
• @Abrahamlure that's correct when $\alpha\le2$ because the limit exists anyways when $\alpha>0$ your method is the same – RE60K Mar 24 '15 at 18:35