# Limit of one to the power of infinity

• Avichal
In summary, for the given problem, the limit x->0 (f(x))g(x) can be written as e(f(x) - 1) * g(x), using equation (1) and the fact that (1+u) = [(1+u)^{1/u}]^u. This can also be written as e^{g\ln(f)}=e^{\frac{\ln(f)}{\frac{1}{g}}}, where the main fraction in the exponent is of L'Hopital niceness. Regarding the general limits doubt, the value of the limit depends on the ratio of the rates at which the variables approach zero. In the case of (1 + x)1/x, the ratio

## 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 elimit x->0 (f(x) - 1) * g(x)
But I don't know how to proceed. I have to somehow use equation (1) but how?

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

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)

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!

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?

"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

arildno said:
"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!

arildno said:
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 $$e^{\frac{\ln(f)}{\frac{1}{g}}} = e^{f-1*g}$$

## 1. What is the limit of one to the power of infinity?

The limit of one to the power of infinity is undefined. This means that as the value of the exponent (infinity) increases, the value of one will approach infinity but never reach it.

## 2. Can the limit of one to the power of infinity be solved mathematically?

No, the limit of one to the power of infinity cannot be solved mathematically because it is an indeterminate form. This means that the limit cannot be determined solely based on the given information.

## 3. How is the limit of one to the power of infinity used in real-world applications?

The limit of one to the power of infinity is often used in calculus and other mathematical concepts to describe the behavior of functions as their input approaches infinity. It is also used in physics and engineering to model the behavior of systems with changing variables.

## 4. What is the difference between the limit of one to the power of infinity and the limit of one divided by zero?

The limit of one to the power of infinity is an indeterminate form, while the limit of one divided by zero is an undefined form. This means that the first limit cannot be solved mathematically, while the second limit does not exist at all.

## 5. Can the limit of one to the power of infinity have a specific value?

No, the limit of one to the power of infinity cannot have a specific value. As mentioned earlier, this limit is undefined and cannot be determined without additional information or context.