1. ## Limit of function

Hi everyone!

First of all thank you to all of you who will answer.

I am attending Calculus course at the university.

I found this limit particularly hard for me to compute:

Limit f x→inf = ?

And f(x) = (3^x+7^x)^(1/x)

I can only get to take it to an exponential form.

I send it to an online limit solver and the answer was: 7.

But I cannot past an indefinite inf^0

Thank you very much.

2. Originally Posted by coltelino
Hi everyone!

First of all thank you to all of you who will answer.

I am attending Calculus course at the university.

I found this limit particularly hard for me to compute:

Limit f x→inf = ?

And f(x) = (3^x+7^x)^(1/x)

I can only get to take it to an exponential form.

I send it to an online limit solver and the answer was: 7.

But I cannot past an indefinite inf^0

Thank you very much.
use logs, or rewrite the function.

Notice that $\displaystyle (3^x + 7^x)^{1/x} = \exp (\ln [(3^x + 7^x)^{1/x}]) = \exp \left( \frac 1x \cdot \ln (3^x + 7^x) \right) = \exp \left( \frac {\ln (3^x + 7^x)}x\right)$

Now take the limit, apply L'Hopital's rule to finish up

3. As x approaches infinity,

$\displaystyle 7^x$ becomes infinitely greater than $\displaystyle 3^x$

Therefore the limit is $\displaystyle (7^x)^{\frac{1}{x}}=7^{\frac{x}{x}}=7^1$

4. Here is another way to see the problem.
$\displaystyle 7 = \sqrt[n]{{7^n }} \leqslant \sqrt[n]{{3^n + 7^n }} \leqslant 7\sqrt[n]{2} \to 7$

5. Very cool, Plato !