6
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Finding Limit

$$\lim_{x \to \infty} (2^x + 3^x + 5^x + 7 ^x + 11 ^x +13^x)^{\frac{1}{x}}$$

So I let

$$y = (2^x + 3^x + 5^x + 7 ^x + 11 ^x +13^x)^{\frac{1}{x}}$$

$\ln$ both sides:

$$\ln{y} = \frac{1}{x} \ln {(2^x + 3^x + 5^x + 7 ^x + 11 ^x +13^x)}$$

Now what?

  • 0
    Have you tried plugging in values of $x$?2012-02-24
  • 0
    Well, your $\ln y$ has indeterminate form $\infty/\infty$2012-02-24
  • 0
    Maybe use the idea [here](http://math.stackexchange.com/questions/111661/is-3x-lt-1-2x-3x-lt-3-cdot-3x-right)?2012-02-24

1 Answers 1

13

From the idea in the question here, for $x>0$: $$ 13^x<2^x+3^x+5^x+7^x+11^x+13^x <6\cdot 13^x; $$ whence $$ 13 <(2^x+3^x+5^x+7^x+11^x+13^x )^{1/x}<6^{1/x}\cdot13. $$ Now use the squeeze theorem.

  • 2
    In general, if $a_i \gt 0$, then $$\lim_{x \to \infty} (a_1^x + a_2^x + \dots + a_n^x)^{\frac{1}{x}} = \max a_i$$. There is also a similar formula for $\min a_i$ (taking $x \to -\infty$).2012-02-24
  • 1
    @Aryabhata That shouldn't be hard to prove. Putting $${a_n}^x < \sum\limits_{k = 1}^n {{a_n}^x} < n{a_n}^x$$ should do the trick.2012-02-24
  • 0
    @PeterT.off: Right, that works.2012-02-24