Question

lim x->0 of (1-ax)^(1/x)

Answer 1

Is this e?

Answer 2

lol, too good to be true

Answer 3

what do you mean by is this e?

Answer 4

I meant the constant e, but I don't want to conclude that, I have a bad feeling about it ;)

Answer 5

the question does not have e in it but i believe the answer will, i think i have to natural log it or something. i really am not sure.

Answer 6

answer is 1

Answer 7

msamido, can you please show the steps?

Answer 8

Uhh, this limit seems like one indeterminate form

Answer 9

yeah. i have to turn it into a fraction and apply the la'hopital rule. but , i dont know how.

Answer 10

\[\huge \lim_{x \rightarrow 0} (1-ax)^{1/x}\]
Indeterminate form of type 1^infinity. Transform using:
\[\huge \lim_{x \rightarrow 0} (1-a x)^{1/x} = e^{\lim_{x \rightarrow 0} \frac{ln(1-a x)}{x}}\]
into:
\[\huge e^{\lim_{x \rightarrow 0} \frac{ln(1-a x)}{x}} \]
Indeterminate form of type 0/0, so use L'Hospital's rule
\[\huge e^{\lim_{x \rightarrow 0} \frac{a}{a x-1}} \]
Factor out constants:
\[\huge e^{a (\lim_{x->0} \frac{1}{a x-1})} \]
The limit of a quotient is the quotient of the limits:
\[\huge e^{\frac{a}{lim_{x->0} (a x-1)}} \]
The limit of a x-1 as x approaches 0 is -1:
\[\huge \frac{1}{e^a}\ = e^{-a}]

Answer 11

Ugh syntax...

The limit of a x-1 as x approaches 0 is -1: \[\huge \frac{1}{e^a}\ = e^{-a}\]

Answer 12

This was kind of a pain to type up with all those {}'s :-P

Are you lost on any step? I don't think I have any more typos, but I've been trying to weed those out for the past few minutes, so hey there could be more

Answer 13

lim x->0 (1-ax)^(1/x)
introducing natural logarithm..
=(1/x)ln(1-ax)
=lim x->o((1/x)(ln(1-ax))
when substituting u will get indeterminant of the form 0/0 them we apply L'HOSPITALS RULE
=limx->0(1/(1-ax))
ax will approach 0 when x -> 0
=1/1
=1

Answer 14

YES. I GOT ITTTTTT! :D thank you sooooooooooo much!

Answer 15

thanks!!