I want to prove this exercise:

Let $x_n \to x$ and $y_n \to y$ for $n \to \infty$ Prove that, $x_n^{y_n} \to x^y$.

My attempt:

Let $\epsilon > 0$ and $N_1 \in \mathbb{N}$ such that $|x_n - x| < \epsilon \forall n \geq N_1 $ and $N_2 \in \mathbb{N}$ such that $|y_n - y| < \epsilon \forall n \geq N_2 $ Then:

$$|x_n^{y_n} - x^y| =x^y |e^{(y_n-y)\ln x_n + y \ln(x_n/x)}-1|$$

What to do now?

  • 2
    $\begingroup$ Should there be an assumption that $x>0$ and $y>0$? $\endgroup$ – user21467 Nov 3 '14 at 21:33

If $x=0$, it's trivial, suppose $x >0$

$$\log \dfrac{x_n^{y_n}}{x^y} = y_n \log x_n - y\log x \to 0$$


$$|y_n \log x_n - y\log x|\leq |y_n - y||\log x_n| + |y |\log x_n - \log x| \to 0$$

since $y_n - y \to 0$, $\log x_n - \log x \to 0$ and $\log x_n$ is bounded.

So $\dfrac{x_n^{y_n}}{x^y} \to 1$ by the continuity of $\log(x)$, then $x_n^{y_n} \to x^y$


Assuming that $x_n>0$ for all n and $x>0$,

$x_n\to x\implies \ln x_n\to \ln x$,

so $y_n\to y\implies y_n\ln x_n\to y\ln x\implies e^{y_n\ln x_n}\to e^{y\ln x}\implies x_n^{y_n}\to x^y$

(since the functions $f(x)=\ln x$ and $g(x)=e^x$ are continuous, and

the limit of the product is the product of the limits).


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