This is the limit I'm trying to solve: $\lim \limits_{x\to1}\left(\frac
x {x-1} - \frac1 {\log(x)}
\right)$
I thought: let's define $x=k+1$, so that $k\to0$ as $x\to1$.
Then it becomes:
$$\lim \limits_{k\to0}\left(\frac
{k+1} {k} - \frac1 {\log(k+1)}
\right)$$
and then,
$$\lim \limits_{k\to0}\left(\frac
{k+1} {k} - \frac1 {\frac {\log(k+1)\times k}k}
\right)=\lim \limits_{k\to0}\left(\frac
{k+1} {k} - \frac1 {k}
\right).$$
Which results in $\frac k k$ , that should be 1, but wolfram says it's $\frac 1 2$...
Did I do something illegal?
Answer
Yes, the illegal part is this step:
$$\lim_{k \to 0}\frac1{\frac{\log(k + 1)}k}\frac1k = \lim_{k \to 0}\frac1k$$
I see that you applied the known limit
$$\lim_{t \to 0}\frac{\log(1 + t)}t = 1$$
but the fact is that
$$\lim_{x \to \alpha}f(x)g(x) = \lim_{x \to \alpha}f(x)\times\lim_{x \to \alpha}g(x)$$
is only valid when both limit are finite. In your case you're left with $\lim\limits_{k \to 0}\frac1k$, which not only is not finite, but does not exist entirely.
If you are looking for a way to evaluate your limit, I'd suggest MacLauring (that is, a Taylor expansion around $x =0$), which is the simplest and most elegant way. But since you said that you can't use Taylor yet, I fear your only possibility is going with L'Hospital.
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