Thread: Help... limits of a log function with exponents

1. Help... limits of a log function with exponents

Hey, I am having trouble with the following problem:

find lim as x -> 1 from the negative side of

1 / (log[base 3](1 - 2^(1 / (x - 1)))

Please show your steps in solving... no matter what i do the exponent
1/(x-1) is killing me.

Thanks,
Landyacht

2. lim as x->1 from the negative side means that x is a little less than 1.

(x-1) is a very small negative number.

1/(x-1) is 1/(a very small negative number), which is a very big negative number

2^(1/(x-1)) is 2^(a big negative number), which is a small positive number (exponential graph 2^x has horizontal asymptote of 0).

1-2^(1/(x-1)) is just 1-a small positive number, or a number just less than 1.

log[base 3](1-2^(1/(x-1))) is just log[base 3](a number just less than 1), which is a very small negative number (log(something < 1) is negative).

1/log[base 3](1-2^(1/(x-1))) is just 1/a very small negative number, which is a large negative number.

if x->1 and gets infinitely close to 1 from the left, then 1/log[base 3](1-2^(1/(x-1))) approaches negative infinity.

3. Thanks comssa.

This may be a stupid question but...:
Is there anyway to find the solution by rearranging though to make the function defined at x = 1, therefore allowing you to find the limit?

Like a lot of other questions I have done.

4. I am not sure. If the function can be arranged so that it is defined at x=1, then f(1) would be weird. At x=1, if you approach from the left, you get closer to negative infinity. I think the graph is undefined as you approach x=1 from the right because you can't have log of a negative number and get a real answer. So if you were to somehow arrange it so that f(1) exists, what would f(1) be?

Also, your question asks for the limit as x->1 from the negative side, so finding f(1) would not be helpful if the limit as x->1 from the positive side is different, which it is.

Hope that helps!