In my country education system, in Math we don't have the right to use L'Hopital's rule to solve indeterminate forms limits for this year.
Instead we use differnt techniques, such as expanding expressions and recently we used change of variable in order to compute limits for different functions that have $ln$ or $e^x$ in them .
For $\ln$ we know the following limits, we don't need to prove them :
$$\lim_{x\to\infty} \ln x/x = 0^+$$
$$\lim_{x\to 0}\ln (x+1)/x = 1 $$
$$\lim_{x\to\0} \ln (x)x = 0^-$$
Then, whenever we have a limit with ln that yield an indeterminate form we use change of variable in order to get a limit that is similar to one of these 3 limits .
Example:
For example we need to compute the following limit :
$$\lim_{x\to-\infty} \frac{x}{\ln|x|}$$
For this one we get an indeterminate form, I need to use the change of variable to try to get a limit that is similar to one of the previous 3 limits, let :
$t = -x <-> x = -t$ then we would have the following :
$$\lim_{t\to\infty} \frac{-t}{\ln(t)}$$
$$\lim_{t\to\infty} \frac{-1}{\frac{\ln(t)}{t}} = -\infty$$
As you can see in the denominator we got the expression as the first limit expression, then I can easily solve it .
The Problem:
My problem, is that I'm really bad at changing variable, I can't find which one would suite the problem as I have very weak algebra skills .
Here are some limits that I couldn't find a way to solve using this method :
$$\lim_{x\to0^+} \frac{\ln(x+1)}{\sqrt{x}} $$
$$\lim_{x\to\infty} \frac{\ln(x)^2}{x}$$
So How Can I solve such limits using the method I described ?
How Can I Find the right way to change a variable for a given complicated limit ? Is there a way to follow ?
Note: As noted previously, I can't use L'Hospital's Rule for this or any other technique except changing variables .
Best Answer
For the first limit, I think you break the limit up like this: $$\lim_{x \to 0} \frac{\ln(x+1)}{\sqrt x}=\lim_{x \to 0} \sqrt x\frac{\ln(x+1)}{x}$$ We know that as $x \to 0$, we have $\sqrt x \to 0$ and from your given limits, $\frac{\ln(x+1)}{x} \to 1$, so we get $0\cdot 1=0$.
For the second limit, I think we should do $x=t^2$ to get $x^2$ on the bottom. That way, there is a square on both the numerator and denominator: $$\lim_{t \to \infty} \frac{\ln^2(t^2)}{t^2}=\frac{(2\ln(t))^2}{t^2}=4\left(\frac{\ln(t)}{t}\right)^2$$ By your given limits, we have $\frac{\ln(t)}{t} \to 0$, so by squaring that and multiplying by $4$, we get the limit's value to be $0$.