Finding limit of the following function

Question

I need to find $$\lim_{x\to a}(\log_ax)^{1/(x-a)}$$ Isnt the answer simply 1 as the logarithm inside is actually 1 and not tending to 1 to make it an indeterminant form.

Answer 1

When you compute a limit, say

$$\lim_{x\to a}f(x),$$

all values of $x$ in a neighborhood of $a$ are considered. But not $x=a$. Hence you can't have $\log_ax=1$.

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By the way, if you assume that $\log_ax=1$, for coherence you must also admit that $x-a=0$ and the expression under the limit would be $1^{1/0}$, which does not make sense.

Of course, it is true that

$$\lim_{x\to a}(\log_aa)^{1/(x-a)}=1$$

but this is a quite different question.

Answer 2

Set $x-a=ay, x=a(1+y),\log_ax=1+\log_a(1+y)$

$$\lim_{x\to a}(\log_ax)^{1/(x-a)}=\lim_{y\to0}(1+\log_a(1+y))^{1/ay}$$

$$=\left(\lim_{y\to0}(1+\log_a(1+y))^{1/\log_a(1+y)}\right)^{\lim_{y\to0}\dfrac{\log_a(1+y)}{ay}}$$

The inner limit converges to $e$

$$\lim_{y\to0}\dfrac{\log_a(1+y)}{ay}=\dfrac{\log_ae}a\lim_{y\to0}\dfrac{\ln(1+y)}y=?$$

Answer 3

Assuming $a>0,a\ne 1$, we get: $$\lim_{x\to a}(\log_ax)^{1/(x-a)}=\lim_{x\to a}\left[(1+\log_ax-1)^{\frac{1}{\log_a x-1}}\right]^{\frac{\log_a x-1}{x-a}}=\\ e^{\lim_\limits{x\to a} \frac{\log_a x-1}{x-a}}\stackrel{LR}{=}e^{\lim_\limits{x\to a}\frac{1}{x\ln a}}=e^{\frac1{a\ln a}}.$$