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Someone can explain how can I resolve this limit please?

$$ \lim_{x \to \infty} (2^x + 3^x)^{1/x} $$

I tried to convert to exponential

$$ \lim_{x \to \infty} \exp \left(\tfrac{1}{x} \ln(2^{x} + 3^{x})\right) $$

$$ \exp \left( \lim_{x \to \infty} \tfrac{1}{x}\ln(2^{x} + 3^{x}) \right) $$

In this part I think applied L'Hospital $$ \exp \left( \lim_{x \to \infty} \frac{\ln(2^{x} + 3^{x})}{x} \right) $$

I noticed that I can't get to answer

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Seems interesting to me. If you've tried anything yet, post the work here, so that we can analyze your work and can tell you about the mistakes in your method. –  Kushashwa Ravi Shrimali Jul 3 at 0:00
    
How about taking $ln$ of $\left(2^x + 3^x \right)^{1/x}$ –  Ahmed Jul 3 at 0:03
    
@user17629 : Please prefer to use \rightarrow instead of -> in TeX ! –  Kushashwa Ravi Shrimali Jul 3 at 0:08

3 Answers 3

up vote 8 down vote accepted

Let $x$ be positive. Note that $3^x\le 2^x+3^x\le 2\cdot 3^x$.

Thus $$3\le (2^x+3^x)^{1/x}\le 3\cdot 2^{1/x}.$$

Since $2^{1/x}\to 1$ as $x\to\infty$, the result follows by Squeezing.

Remark: Your more complicated method will work, if we use the fact that $\ln(2^x+3^x)=\ln[(3^x)(1+(2/3)^x]=x\ln 3+\ln(1+(2/3)^x)$. Now divide by $x$ and take the limit.

Alternately, one can work with your expression, and evaluate $\lim_{x\to\infty}\frac{\ln(2^x+3^x)}{x}$ using L'Hospital's Rule.

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This is just the fact that the $p$-norms converge to the max-norm. –  lhf Jul 3 at 3:33

$$(2^x+3^x)^{\frac{1}{x}}=3\left(1+\left(\frac{2}{3}\right)^x\right)^{\frac{1}{x}}$$

Now $$1 \leq \left(1+\left(\frac{2}{3}\right)^x\right)^{\frac{1}{x}} \leq 1+\left(\frac{2}{3}\right)^x $$

So $$\left(1+\left(\frac{2}{3}\right)^x\right)^{\frac{1}{x}}\rightarrow 1$$

and $$(2^x+3^x)^{\frac{1}{x}}\rightarrow 3$$

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Hint: Factor $3^x$ inside the parenthesis.

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