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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.


marked as duplicate by Guy Fsone, Parcly Taxel, Hans Engler, Chris Godsil, M. Vinay Jan 28 '18 at 14:19

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  • $\begingroup$ is that$\ \lim_{n \to \infty} (3^n+7^n)^{1/n}$ ? $\endgroup$ – CTSnake Apr 28 '17 at 5:43
  • $\begingroup$ @CTSnake sorry that was a typo. $\endgroup$ – Navin Apr 28 '17 at 5:48
  • 2
    $\begingroup$ This was asked tons of times already. $\endgroup$ – Did Apr 28 '17 at 5:49
  • $\begingroup$ See this and its related posts. $\endgroup$ – StubbornAtom Apr 28 '17 at 7:03


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.


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


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


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