Find the limit of $(7^n + 9^n)^{(1/n)}$ when n goes to $\infty$

Question

I tried this using L' Hopitals rule. But I always get a limit that is not defined.

$$\lim_{n \to\infty}(7^n +9^n)^{(1/n)}$$

Let $$y = (7^n +9^n)^{(1/n)}$$
then take the $log$ of both side,
$$ln(y)= ln((7^n +9^n)^{(1/n)})$$
$$ln(y)= (1/n) * ln(7^n +9^n)$$
$$ln(y) = (ln(7^n +9^n))/n$$

then we find the limit of both side when n goes to $\infty$
$$\lim_{n \to\infty}(ln(y))= \lim_{n \to\infty}((ln(7^n +9^n))/n)$$

we can see the limit of numerator and the denominator is infinity. So, we apply L'Hopitals rule,

$$\lim_{n\to\infty}(ln(y)) = (1/(7^n +9^n))*(7^n*ln7+ 9^n*ln9) = (7^n*ln7+ 9^n*ln9)/(7^n +9^n)$$

Again this limit is not defined as we get $\infty$ by $\infty$
If I apply L'Hopitals again and the limit of the result will be the same.

Can someone help me to find the answer by L'Hopitals or an Alternative way.

Answer 1

$$\sqrt[n]{9^n+7^n}=9\sqrt[n]{1+\left(\frac{7}{9}\right)^n}\rightarrow9.$$