# How to evaluate $\lim_{x\rightarrow +\infty } \sqrt{a^x+b^x} = ?$

If a>0 and b>0,
$\lim_{x\rightarrow +\infty } \sqrt[x]{a^x+b^x} = ?$

What I was trying to do:
Suppose a>b. Then, for sufficiently large values of x, $a^x >> b^x$; so $\sqrt[x]{a^x+b^x} \rightarrow \sqrt[x]{a^x} \rightarrow a$ when $x \rightarrow +\infty$.

Is that idea correct? How can I formalize it?

#### Solutions Collecting From Web of "How to evaluate $\lim_{x\rightarrow +\infty } \sqrt{a^x+b^x} = ?$"

Assume $a>b$
$$a^x\leq a^x+b^x\leq 2a^x$$

$$(a^x)^{\frac{1}{x}}\leq (a^x+b^x)^{\frac{1}{x}}\leq (2a^x)^{\frac{1}{x}}$$

$$a\leq (a^x+b^x)^{\frac{1}{x}}\leq 2^{\frac{1}{x}}a$$

Hint : If $a>b$, consider $a^x+b^x=a^x(1+(\frac{b}{a})^x)$ and use the fact that
the expression in paranthesis tends to $1$.

You can go this way. Assuming $b<a$

$$(a^x+b^x)^{1/x} = a \,e^{\frac{1}{x}\ln(1+(b/a)^x) }= a \,e^{\frac{1}{x}((b/a)^x-(b/a)^{2x}/2+\dots ) }\longrightarrow _{x\to \infty } a .$$

I think you can finish it!

Note: You need the following power series

$$\ln(1+t) = t-\frac{t^2}{2}+ \frac{t^3}{3}-\dots\,.$$