How do you find the limit of #xlnx# as #x-0^-#?

Question

How do you find the limit of #xlnx# as #x->0^-#?

Charles Arocha · Accepted Answer

There is no limit as #x# approaches #0# from below since #ln x# is undefined for negative numbers. Instead, I will demonstrate how to find the right-handed limit, i.e., as #x->0^+#.

Here is a graph:

So we should expect the answer to be zero. Now, to do this we can't use the product rule, since the limit of #ln x# diverges as #x->0^+#, we have to be more clever.

L'HOPITAL'S RULE: You can Google the precise formulation of this, and the conditions where it applies, but roughly speaking, the rule states that if you have a limit of the form #\infty/\infty# or #0/0#, then you can differentiate both parts to evaluate the limit. We need to rewrite the question to do this:

#lim_{x->0^+}x ln x=lim_{x->0^+}ln x / {1/x}=lim_{x->0^+}-{1/x}/{1/x^2}=lim_{x->0^+}-x=0.#

You could probably figure out other ways to evaluate this limit, maybe using the squeeze theorem with upper bound #x^2# and something else for your lower bound, but L'Hopital's rule is how everyone would evaluate this limit.

Samantha Adkison · Answer

undefined To find the limit of xlnx as x approaches 0 from the left (x->0^-), we can use L'Hôpital's Rule.

First, we rewrite the expression as ln(x)/1/x.

Next, we differentiate the numerator and denominator separately. The derivative of ln(x) is 1/x, and the derivative of 1/x is -1/x^2.

Now, we evaluate the limit of the derivatives as x approaches 0 from the left.

The limit of 1/x as x approaches 0 from the left is -∞, and the limit of -1/x^2 as x approaches 0 from the left is also -∞.

Since both limits are -∞, we can apply L'Hôpital's Rule again.

Differentiating the numerator and denominator once more, we get -1/x and 2/x^3, respectively.

Taking the limit of these derivatives as x approaches 0 from the left, we find that both limits are -∞.

Since both limits are -∞, we can continue applying L'Hôpital's Rule until we reach a determinate form.

After applying L'Hôpital's Rule multiple times, we find that the limit of xlnx as x approaches 0 from the left is 0.

How do you find the limit of #xlnx# as #x-&gt;0^-#?

How do you find the limit of #xlnx# as #x->0^-#?