Limit of one to the power of infinity

[Avichal]

Homework Statement

Find limit _x->0 (f(x))^g(x).
Here limit _x->0 f(x) = 1 and limit _x->0 g(x) = ∞

Homework Equations

I know that limit _x->0 (1 + x)^1/x = e ... (1)

The Attempt at a Solution

I have the answer which is e^{limit _x->0 (f(x) - 1) * g(x)}
But I don't know how to proceed. I have to somehow use equation (1) but how?

 

[ppham27]

Try writing $$f(x) = 1 + [f(x) - 1].$$ Also, note that $$(1+u) = [(1+u)^{1/u}]^u.$$

 

[Avichal]

Okay I got some idea. Is this right?

f(x)^g(x) = ((1 + f(x) -1)^{1/(f(x) - 1)})^{(f(x) - 1) * g(x)}
Since (1 + f(x) -1)^{1/(f(x) - 1)} = e ... using (1)
∴ ans = e^{(f(x) - 1) * g(x)}

 

[arildno]

Note that this can be written as:
$$e^{g\ln(f)}=e^{\frac{\ln(f)}{\frac{1}{g}}}$$
The main fraction in the exponent is of L'Hopital niceness.
In theory, that is!

 

[Avichal]

Now that I asked this question I have another related doubt.
This isn't really related to the original question but a general limits doubt.

We know that (1 + x)^1/x = e as x → 0
But why isn't (1 + x²)^1/x = e as x → 0

Does the fact that x² approaches 0 faster than x change the value of the limit? But if the power was also x² instead of x then the limit would be e. How come?

 

[arildno]

"But if the power was also x2 instead of x then the limit would be e. How come?"
Because it is the ratio of the RATES by which they tend to zero that is critical for what the value will be.

Look at (1+2x)^(1/x), for example. Here, x goes to zero twice as fast than 2x, meaning the limit is e^(2), rather than e

 

[Avichal]

"But if the power was also x2 instead of x then the limit would be e. How come?"
Because it is the ratio of the RATES by which they tend to zero that is critical for what the value will be.

Look at (1+2x)^(1/x), for example. Here, x goes to zero twice as fast than 2x, meaning the limit is e^(2), rather than e

Oh ok, yes. Thank you!

Note that this can be written as:
$$e^{g\ln(f)}=e^{\frac{\ln(f)}{\frac{1}{g}}}$$
The main fraction in the exponent is of L'Hopital niceness.
In theory, that is!

I didn't understand the R.H.S.
Is $$e^{\frac{\ln(f)}{\frac{1}{g}}} = e^{f-1*g}$$

 

[arildno]

I didn't read your question properly, so unfortunately, I gave a wholly irrelevant answer. Sorry about that..