Evaluating $\lim_{x \to \infty} (3^x+7^x)^{1/x}$

Question

The question is to evaluate $$\lim_{x \to \infty} (3^x+7^x)^{1/x}$$

I tried evaluating the limit as $exp(\lim_{x\to \infty} (3^x+7^x-1)(1/x))$.I couldn't proceed after this.Any help shall be highly appreciated. Thanks.

is that$\ \lim_{n \to \infty} (3^n+7^n)^{1/n}$ ? – CTSnake Apr 28 '17 at 05:43 — CTSnake, Apr 28 '17 at 05:43
@CTSnake sorry that was a typo. – Navin Apr 28 '17 at 05:48 — Navin, Apr 28 '17 at 05:48
This was asked tons of times already. – Did Apr 28 '17 at 05:49 — Did, Apr 28 '17 at 05:49
See this and its related posts. – StubbornAtom Apr 28 '17 at 07:03 — StubbornAtom, Apr 28 '17 at 07:03

mathworker21 · Answer 1 · 2017-04-28T06:51:33.653

6

$7 = (7^x)^{1/x} \le (3^x+7^x)^{1/x} \le (7^x+7^x)^{1/x} = 2^{1/x}7$, so the limit is $7$.

edited Apr 28 '17 at 06:51

answered Apr 28 '17 at 05:49

mathworker21

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score 2 · Answer 2 · edited Apr 28 '17 at 06:00

2

Well, $7 < (3^x+7^x)^{1/x} = 7((\frac{3}{7})^x + 1)^{1/x} < 7(2)^{1/x} \rightarrow 7$ when $x\to \infty$.

edited Apr 28 '17 at 06:00

user1551

149,263

answered Apr 28 '17 at 05:48

GAVD

7,436

score 1 · Accepted Answer · answered Apr 28 '17 at 05:54

1

HINT:

You're on the right track - almost.

If you wanted to keep going the way you are consider,

$$e^{lim_{x \to \infty}\bigg(\frac{ln(3^x + 7^x)}{x}\bigg)}$$

Next, you can apply logarithm laws again, and then l'Hopital's rule.

answered Apr 28 '17 at 05:54

Retty

497

Evaluating $\lim_{x \to \infty} (3^x+7^x)^{1/x}$

3 Answers3