Calculate the limit: $\lim_{x\to\infty} \sqrt[x]{3^x+7^x}$

Question

Calculate the limit: $$\lim_{x\to\infty} \sqrt[x]{3^x+7^x}$$

I'm pretty much clueless on how to approach this. I've tried using the identity of $c^x = e^{x \cdot \ln(c)}$ but that led me to nothing. Also I've tried replacing $x$ with $t=\frac{1}{x}$ such that I would end up with $\lim_{t\to 0} (3^{1/t} + 7^{1/t})^{1/t}$ however I've reached yet again a dead end.

Any suggestions or even hints on what should I do next?

Answer 1

Note that

$$\sqrt[x]{3^x+7^x}=7\sqrt[x]{1+(3/7)^x}=7\cdot \large{e^{\frac{\log{1+(3/7)^x}}{x}}}\to7$$

Answer 2

Note that this is the same as

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

and $|3/7| < 1$. What can you conclude?

Answer 3

\begin{align*} \log(3^{x}+7^{x})^{1/x}=\log 7\left(\left(\dfrac{3}{7}\right)^{x}+1\right)^{1/x}=\log 7+\dfrac{\log((3/7)^{x}+1)}{x}=\cdots \end{align*}

Perhaps, \begin{align*} 1\leq\left((3/7)^{x}+1\right)^{1/x}\leq(1+1)^{1/x}=2^{1/x}, \end{align*} so \begin{align*} 7\leq(3^{x}+7^{x})^{1/x}=7((3/7)^{x}+1)^{1/x}\leq 7\cdot 2^{1/x}, \end{align*} and $7\cdot 2^{1/x}\rightarrow 7$, then Squeeze Theorem concludes.

Answer 4

it is $$7\cdot \sqrt[x]{\left(\frac{3}{7}\right)^x+1}$$

Answer 5

$\sqrt[x]{7^x}\le \sqrt[x]{3^x +7^x}\le \sqrt[x]{2\cdot 7^x}$. Now squeeze.

Answer 6

Note:

Let $x>0.$

$ 7 \lt 7(1+(3/7)^x)^{1/x} \lt 7(2)^{1/x}$.

And now consider $x \rightarrow \infty$.