Show that $\lim x^{a_n} = \lim x^{b_n}$. Let $x>0$ and $\lim a_n = \lim b_n$. Show that $\lim x^{a_n} = \lim x^{b_n}$. 
So, basic, this is the question. I already tried to see what I know about it, and since the limit of $a_n$ and $b_n$ exist, $a_n$ and $b_n$ are the Cauchy type and limited. I started trying with $\lvert\,x^{a_n}-x^{b_n}\rvert = \lvert\,x^{a_n}(1 -x^{b_n-a_n})\rvert= \lvert\,x^{a_n}\rvert\,\lvert\,(1 -x^{b_n-a_n})\rvert$ and since $a_n$ is limited, we know that $\lvert\,a_n\rvert<L$ for any $L.$ So $\lvert\,x^{a_n}\rvert\,\lvert(1 -x^{b_n-a_n})\rvert \leq \lvert\,x^L\rvert\,\lvert(1 -x^{b_n-a_n})\rvert$ and I've already shown in my calculations that $a_n$ and $b_n$ are equivalents, but I didn't continue after that. 
 A: You are proceeding in the right direction. The problem is equivalent to showing that if some sequence $d_{n} $ tends to $0$ then $x^{d_{n}} \to 1$. This is not difficult and based on the following simple limit $x^{1/n}\to 1$. An easy proof of this is via a special case of Cesaro-Stolz:

If $a_{n} $ is positive and $a_{n+1}/a_{n}\to L$ then $a_{n} ^{1/n}\to L$. 

Putting $a_{n} =x$ we easily get $x^{1/n}\to 1$. Taking reciprocals we see that $x^{-1/n}\to 1$. Therefore given any $\epsilon>0$ there is a positive integer $N$ such that $$|x^{1/n}-1|<\epsilon,|x^{-1/n}-1|<\epsilon\tag{1}$$ for $n\geq N$. Next $d_{n} \to 0$ so there is a positive integer $M$ such that $$-\frac{1}{N}<d_{n}<\frac{1}{N}\tag{2}$$ for all $n\geq M$. If $n\geq M$ then by the above inequality $x^{d_{n}} $ lies between $x^{-1/N}$ and $x^{1/N}$. It now follows from $(1)$ that $$|x^{d_{n}} - 1|<\epsilon$$ for all $n\geq M$. Thus $x^{d_{n}} \to 1$. Now put $d_{n} =a_{n} - b_{n} $ for your problem.
The above approach avoids use of any special functions like $\log$ or $\exp$ and instead relies on basic algebra.

The above is taken from my blog post and it is a key step in developing a theory of general power $x^{a} $ for real $a, x$.  The approach uses sequences of rationals $a_{n}\to a$ and defines $x^{a} =\lim_{n\to\infty} x^{a_{n}} $.
A: Nice starting point! Now, observe that$$|1-x^{b_n-a_n}|=\bigl|1-e^{(b_n-a_n)\log x}\bigr|\leqslant2\bigl|(b_n-a_n)\log x\bigr|$$if $b_n-a_n$ is close enough to $0$, because$$\lim_{x\to0}\frac{e^x-1}x=1<2.$$Now, since $\lim_{n\to\infty}b_n-a_n=0$, the hypothesis “$b_n-a_n$ is close enough to $0$” holds if $n$ is large enough.
A: You are on the right track as i see from your attempt.
If you are familiar with the sequential characterization of continuity then:
Let $c_n=b_n-a_n \to 0$
Thus $x^a$ is a continuous function for $x>0$ thus $x^{c_n} \to x^0=1$
Therefore $$1-x^{c_n} \to 1-1=0$$
Using this you get the result.
A: Just take the logarithm since it is continuous you have
$$\log x \lim  a_n= \lim ( a_n\log x ) = \lim  \log ( x^{a_n}) = \log \lim x^{a_n} =\log \lim x^{b_n} = \log x \lim  b_n  $$
if $\log x \ne 0$ you can divide by it and get the result. You also see that you need to exclude the case $x=1$.
A: Another proof: If $\lim_n a_n = \lim_n b_n$, then $\lim_n (a_n / \ln x) = \lim_n (b_n / \ln x)$. Now if you know $e^t$ is differentiable, then you know it's also continuous, so $\lim_n e^{a_n / \ln x} = \lim_n e^{b_n / \ln x}$. But that just means $\lim_n x^{a_n} = \lim_n x^{b_n}$.
