Question

L'hospital's rule and other forms! help please! lim of (1-2x)^1/x as x approaches 0

Answer 1

\[\lim_{x \rightarrow 0} (1-2x)^{1/x}\]

Answer 2

@zepdrix

Answer 3

I can't solve this without making it intricate.
@satellite73

Answer 4

Mmm ya this one is a little tough :o we have to use the ole exponent/log trick.

So we first recognize that the function is approaching the indeterminate form \(\Large 1^{\infty}\) as x approaches 0. So we'll have to try to do something fancy to see what's really going on.

Recall that since the exponential and log functions are inverses of one another we can write something like this:\[\Large u\quad=\quad e^{\ln u}\]
^ That's the kind of thing we want to do with this limit.
So we'll write it like this:\[\LARGE e^{\ln\left[\lim_{x \rightarrow 0} (1-2x)^{1/x}\right]}\]With all of that stuff in the log in the exponent.

Answer 5

Ignore the base exponential right now, just look at the log portion.
We'll pass the (lim) outside of the log.\[\Large \lim_{x\to0}\ln(1-2x)^{1/x}\]

Answer 6

Yeah, that's what I was trying to avoid doing. Guess there's no way out.

Answer 7

Ya I don't know of another method D:

Answer 8

Rule of logs allows us to write it as:\[\Large \lim_{x\to0}\frac{1}{x}\cdot \ln(1-2x)\quad =\quad \lim_{x\to0}\frac{\ln(1-2x)}{x}\]If we evaluate the limit from here we'll see that we're getting the indeterminate form 0/0.
So we FINALLY have it in a form where we can apply L'Hopital!

Answer 9

Bahh, you're not even online XD lol

Answer 10

You could forgo the use of L'Hopital's with a substitution, \(t=\dfrac{1}{x}\), so that \(t\to\infty\) as \(x\to0^+\). Then, use the known limit
\[\lim_{t\to\infty}\left(1+\frac{a}{t}\right)^t=e^{a}\]

Answer 11

@SithsAndGiggles *known. Not by everyone.