Solving limit without L'Hopital I'd like some help in solving this limit without using L'Hopital.
$$\lim_{x\to -\infty}\frac{\ln(1-2x)}{1-\sqrt{1-x}}$$
I've also solved it changing the variable to $y=\sqrt{1-x}$ but I would like to see if there is some other way to solve it because this variable change is not very intuitive IMO.
 A: First try to have a grasp of what's going on here: we're interested in the limit as $x\to-\infty$ of an expression.
This expression involves $\ln(1-2x)$ that is well-defined in a neighborhood of $-\infty$; it approaches $+\infty$ as $x\to-\infty$, but not too quickly, because of the $\ln$.
We also have $1-\sqrt{1-x}$ which is well-defined in a neighborhood of $-\infty$, and approaches $-\infty$ as $x$ approaches $-\infty$, and does so faster (whatever that means) that the $\ln$ part.
This should give you a hint that your expression approaches $0$, by negative values, as $x\to-\infty$.
Now you can express the fact that $\ln(x)$ approaches $+\infty$ as $x\to+\infty$ not too quickly, at least with respect to powers of $x$, by stating:
$$\lim_{x\to+\infty}\frac{\ln(x)}x=0,$$
but in fact, you can solve your problem by using (appropriate algebraic manipulation and) the weaker following fact:
$$\forall x>0,\ \ln(x)<x.$$
This fact is rather easy to understand graphically; its proof depends on your definition of $\ln$.
Now write, for $x<0$ (so that $1-2x>0$):
$$0<\ln(1-2x)=3\frac{\ln\bigl((1-2x)^{1/3}\bigr)}{(1-2x)^{1/3}}(1-2x)^{1/3}<3(1-2x)^{1/3}$$
and then, we obtain for $x<0$ (so that $1-\sqrt{1-x}<0$):
$$0>\frac{\ln(1-2x)}{1-\sqrt{1-x}}>\frac{3(1-2x)^{1/3}}{1-\sqrt{1-x}}.$$
At this point we got rid of the transcendental function $\ln$, and only have algebraic terms. Of course our goal is to conclude using the Squeeze Theorem. So let's rewrite the right-hand side (without the harmless $3$): for all $x<0$,
$$\frac{(1-2x)^{1/3}}{1-\sqrt{1-x}}=\frac{\sqrt{-x}\left(\dfrac1{-x^{3/2}}-\dfrac2{-x^{1/2}}\right)^{1/3}}{\sqrt{-x}\left(\dfrac1{\sqrt{-x}}-\sqrt{\dfrac1{-x}+1}\right)}\underset{x\to-\infty}\longrightarrow\frac0{-1}=0.$$
Conclude with the Squeeze Theorem.
There are many possible variations on this theme. The crucial part is to deal with the $\ln$ appropriately; here I chose to exploit $\ln x<x$ (which is sufficient to solve many limits involving $\ln$).
A: Allright. In the numerator we have a plain $ln$ term. In the denominator we have a squareroot term, which, technically is degree "one half" (OK, don't quote me on the degree definition here) So I learned that the $ln$ is going to lose against any radical and so the limit is zero.
A: Hint: First, amplify with the conjugate, $1+\sqrt{1-x}$. Then let $(1-2x)_{x\to-\infty}=(e^t)_{t\to\infty}$. I trust that things will pretty fast become self-evident.
