Show $log{a_n}rightarrowlog{a}$ as ${nrightarrowinfty}$

Question

Let $(a_n)_{ninmathbb{N}}inmathbb{R}$ and assume $a_n>0$ and $a_nrightarrow a>0$ as $nrightarrowinfty$.
The sequence $a_n$ clearly converges and I am inclined to think that if $a_n$ is convergent with the above statement being true, then
$lim_{nrightarrowinfty}{log(a_n)=log(lim_{nrightarrowinfty}{a_n})}$ but I am not sure how to prove it.

PierreCarre · Accepted Answer

The result can come directly from the continuity of $log$... But, just for using something different, Lagrange's theorem allows you to establish that
$$
|log a_n - log a| = frac{1}{xi_n}|a_n -a|, quad xi_n in (a_n, a)
$$
From the previous equality, we get
$$
lim |log a_n -log a| = lim frac{1}{xi_n}|a_n-a| = frac 1a cdot 0=0,
$$
which in turn means that $lim a_n =a$.

Show $log{a_n}rightarrowlog{a}$ as ${nrightarrowinfty}$

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