$\lim a^{b_n}=a^b$ if $\lim b_n = b$

Question

Let $a \in \mathbb{R}$ and $\lim b_n = b$ (for a real sequence $b_n$). I am trying to show that $\lim a^{b_n}=a^b$ by using the definition of the convergence of a sequence, but I cannot see how to estimate $|a^{b_n} - a^b|$ against $|b_n - b|$ properly.

Answer 1

Hint: $a>0$, $a^{b_n}=\exp \left(b_n\log \left(a\right)\right)$. The function $f\left(x\right)=\exp \left(x\log\left(a\right)\right)$ is continue.

Answer 2

Let $\epsilon > 0$. Let $f(x) = a^x$. I'm assuming $a$ is positive. We want to find an $N$ such that if $n \geq N$, $|a^{b_n}-a^b| < \epsilon$.

Note that $V = (\log_a(a^b-\epsilon),\log_a(a^b+\epsilon))$ is an open interval containing $b$ (because the logarithm is order preserving, this is the same as saying that $f(b) \in f(V) = (a^b - \epsilon, a^b + \epsilon)$.

We can find an $N$ such that $n \geq N$ implies $b_n \in V$. Then $a^{b_n}$ is within $\epsilon$ of $a^b$, as required.

Answer 3

Here is a possibility assuming $a>1$, I'll leave other values up to you.

Given $\varepsilon>0$, write $$\varepsilon'=\frac{\log\bigl(1+\frac\varepsilon{a^b}\bigr)}{\log a}\ .$$ Then $\varepsilon'>0$, so for sufficiently large $n$ we have $$|b_n-b|<\varepsilon'\ .$$ This gives $$a^{b_n}-a^b=a^b(a^{b_n-b}-1)<a^b(a^{\varepsilon'}-1)=\varepsilon$$ and $$a^{b_n}-a^b>a^b(a^{-\varepsilon'}-1) =a^b\left(\frac1{1+\frac\varepsilon{a^b}}-1\right) =-\frac\varepsilon{1+\frac\varepsilon{a^b}} >-\varepsilon\ .$$ So $$|a^{b_n}-a^b|<\varepsilon\ .$$

Answer 4

Herein, we present a way forward that relies only on an elementary pair of inequalities and the squeeze theorem. We begin with a primer.

PRIMER:

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities

$$1+x\le e^x\le \frac{1}{1-x} \tag 1$$

for $x<1$

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First we note that we can write

$$\begin{align} a^{b_n}-a^b&=a^b\left(a^{b_n-b}-1\right)\\\\ &=a^b\left(e^{(b_n-b)\log(a)}-1\right)\\\\ \end{align}$$

We take $n$ large enough to guarantee that $(b_n-b)\log(a)<1$. Then, we have

$$a^b(b_n-b)\log(a) \le a^b\left(e^{(b_n-b)\log(a)}-1\right)\le \frac{a^b(b_n-b)\log(a) }{1-(b_n-b)\log(a) } \tag 2$$

whereupon applying the squeeze theorem to $(2)$ we obtain the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}(a^{b_n}-a^{b})=0}$$

And we are done!

Answer 5

$\{b_n\}$ converges to $b$

$n>N \implies |b_n - b| < \epsilon_1$

$a^{b_n} - a^b = a^b(a^{b_n-b}-1)$

$\{a^{b_n}\}$ converges to $a^b$ if there exists an $\epsilon_1$ such that $|(a^b)(1-a^{\epsilon_1})|<\epsilon$

$\epsilon_1 < \log_a (1-\epsilon a^{-b})$