1

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.

Michael Rozenberg
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ShanOH
  • 554

4 Answers4

8

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

Michael Rozenberg
  • 194,933
3

$$9^{n}<7^{n}+9^{n}<9^{n}+9^{n}=2\cdot 9^{n}$$

$$9<\bigg(7^{n}+9^{n}\bigg)^{\frac{1}{n}}<\bigg(2\cdot 9^{n}\bigg)^{\frac{1}{n}}$$

Using Squeeze Theorem

$$\lim_{n\rightarrow \infty}\bigg(7^{n}+9^{n}\bigg)^{\frac{1}{n}}=9.$$

jacky
  • 5,194
1

In general, for $a>0$ and $b>0$, we have $$\lim_{n\to \infty}\sqrt[n]{a^n+b^n}=\max(a,b)$$ In this case, the given limit is equal to $9$ (just as they answered you above).

azif00
  • 20,792
1

This may be a bit more lengthy, but it is a good general method which is not so widespread.
Using the Cauchy-D'Alembert criterion(see https://math.stackexchange.com/a/3159844/629594), we have
$$\lim \limits _{n\to \infty}\sqrt[n]{7^n+9^n}=\lim \limits _{n\to \infty}\frac{7^{n+1}+9^{n+1}}{7^n+9^n}=\lim \limits _{n\to \infty}\frac{7\cdot\left(\frac{7}{9}\right)^n+9}{\left(\frac{7}{9}\right)^n+1}=9$$

Alexdanut
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