I've got a new problem.
the lim as x ->1 of x/lnx?
when it goes to a number don't you just plug it in but then there's a 0 in the denominator so what now?
Nah you don't just plug it in. That defeats the purpose of a limit.
$\displaystyle \frac{1}{0}$ is the value of the function $\displaystyle \frac{x}{\ln(x)} $ AT the point $\displaystyle x = 1 $. However
$\displaystyle \displaystyle \lim_{x \to 1} \frac{x}{\ln(x)}$ is the value of the function as x APPROACHES 1.
As it happens, the limit you proposed does not exist.
A quick viewing of the graph of the function is all that's needed to show that the limit doesn't exist.
See the graph I've attached.
At x = 1, you can see that the graph isn't defined. As you approach x = 1 from the right, the graph tends towards positive infinity, and as you approach x = 1 from the left, the graph tends towards negative infinity.
The graph does not tend towards a specific value as x tends towards 1. The value it tends towards is undefined, and hence the limit does not exist.
Hi. I'd like to comment, please. I think it's important that beginners realize that as long as the function is continuous at the point in question, simple substitution for limit evaluation is fine. In fact, that should be your first move. However, it's critical that you understand what a limit is. This understanding is cardinal when dealing with discontinuities. We only care about what value F(x) approaches as we near the limit, not what value F(x) yields at the limit. If the function is continuous, just plugging in is fine because they are one in the same.
You're meant to understand that as x --> 1 from the
* right hand side, ln x --> 0 from above (and is therefore small but positive) and hence 1/ln x --> +oo.
* left hand side, ln x --> 0 from below (and is therefore small but negative) and hence 1/ln x --> -oo.
Hence the right hand limit is different to the left hand limit and so the limit doesn't exist.