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Let $\alpha$, $\beta$ be convergent sequences of positive terms with the limits a, b.
Is it true that the sequence $\alpha^\beta$ must also be convergent?

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    HINT: take logarithm. The unique problem arises in the case $a=b=0$. – 2017-02-27
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    Weird notation: if the sequences are $\;\{a_n\}\,,\,\,\{b_n\}\;$ , do you mean $\;\left\{a_n^{b_n}\right\}\;$ ? – 2017-02-27
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    No. $\alpha_n=\frac 1{n^2}=\beta_n$ provides a counterexample (the terms go to $1$). – 2017-02-27

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Yes, it is. Recall that $\ln x$ is a continuous function in $\mathbb R^+$, so $\lim_{n} \ln (a_n)=\ln (\lim_{n} a_n)$ if $a_n$ is a sequence of positive terms.

Now, if $A=\lim_{n} a_n$ and $B=\lim_{n} b_n$

$$\ln(\lim_{n} {a_n}^{b_n})=\lim_{n}(\ln {a_n}^{b_n})=\lim_n [b_n \cdot\ln(a_n)]=\lim_n b_n \cdot \lim_n \ln a_n=B\cdot\ln A = \ln (A^B)$$

and cancelling logarithms, $\lim_{n} {a_n}^{b_n}=A^B$.

Note that this is only true if the expression make sense: for example, $0^0$, $1^\infty$ and $\infty^0$ are not defined, and this rule doesn't apply.

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    what if $A=B=0$? $0^0$ is undefined. – 2017-02-27
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    Well, that is false with $\;a_n=b_n=\frac1n\;$ , since then $$a_n^{b_n}=\frac1{\sqrt[n]n}\xrightarrow[n\to\infty]{}1$$ whereas $\;0^0\;$ isn't even defined... – 2017-02-27
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    We could suppose that $0^0=1,$ as it is defined in some contexts, but consider what happens if $b_n=1/n$ and $a_n=1/n^n$ when $n$ is even, $a_n=1/n$ when $n$ is odd. – 2017-02-27
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    Yes, the rule doesn't work if an expression of the type $0^0$, $1^\infty$ or $\infty^0$ is obtained, as it happens with the quotient rule $$\lim \frac{a_n}{b_n} = \frac{\lim {a_n}}{\lim {b_n}}$$ when numerator and denominator have limit $0$ or $\infty$. – 2017-02-27