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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.

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    $\begingroup$ Genuine curiosity: why do you want to avoid L'Hopital? $\endgroup$
    – Almo
    Commented Mar 20, 2014 at 18:41
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    $\begingroup$ @Almo - see this meta discussion. $\endgroup$
    – 2012ssohn
    Commented Mar 20, 2014 at 18:44
  • $\begingroup$ The limit is $0$. You just need to check that you understand that $\forall \alpha > 0 \frac{ln(x)}{x^{\alpha}} \rightarrow 0$ as $ x \rightarrow \infty$. $\endgroup$
    – Frank
    Commented Mar 20, 2014 at 18:46
  • $\begingroup$ Dood, thanks 2012ssohn! That made a lot of sense to me. :) $\endgroup$
    – Almo
    Commented Mar 20, 2014 at 18:49

3 Answers 3

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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 $\ln$).

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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.

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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.

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