L’Hospital’s Rule and Logarithms

In this post, I’ll use L’Hopital’s Rule (a rule that is important for calculating limits) to calculate the following limit:

$$ \lim_{x \to 0^+} x \ln(x) = 0 $$

This limit uses what is known as the natural logarithm (\( \ln(x) \)), which is very useful in Calculus. Note: \( x \to 0^+ \) signifies that \( x \) approaches \( 0 \) from the right.

Don’t worry if you don’t know the rule. I’ll explain the important parts of that rule as we use it to solve this problem.

First Step

First, notice that we can do this:

$$x = \frac{1}{1/x} $$

Multiplying both sides of this by the natural logarithm \( \ln(x) \) results in:

$$x \ln(x) = \frac{\ln(x)}{1/x} $$

Second Step

Let’s take a look at

$$ \frac{\ln(x)}{1/x} $$

Consider

$$ \lim_{x \to 0^+} \ln(x) $$

and

$$ \lim_{x \to 0^+} \frac{1}{x} $$

If both of these limits are infinite, or if both of these limits are zero, I can use L’Hospital’s Rule. It turns out that

$$ \lim_{x \to 0^+} \ln(x) = -\infty $$

$$ \lim_{x \to 0^+} \frac{1}{x} =\infty $$

So I can use the rule.

Third Step

We can calculate what are known as derivatives of these functions. It turns out that the derivative of \( \ln(x) \) is

$$ \frac{1}{x} $$

and the derivative of \( 1/x \) is

$$ \frac{-1}{x^2} $$

What L’Hospital’s rule says is that

$$ \lim_{x \to 0^+} \frac{\ln(x)}{1/x} = \lim_{x \to 0^+} \frac{\text{derivative of } \ln(x)}{\text{derivative of }1/x} $$

So,

$$ \lim_{x \to 0^+} \frac{\ln(x)}{1/x} = \lim_{x \to 0^+} \frac{\text{derivative of } \ln(x)}{\text{derivative of }1/x} $$

$$ = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} $$

Fourth Step

Notice that we can do this

$$\frac{1/x}{-1/x^2} =\frac{-1/x}{1/x^2} $$

$$ = \frac{-1}{x} \cdot \frac{1}{1/x^2} $$

Also, notice that

$$ \frac{1}{1/x^2} = x^2 $$

So,

$$\frac{1/x}{-1/x^2} = \frac{-1}{x} \cdot x^2$$

$$= \frac{-x^2}{x} = -x $$

Since

$$\frac{1/x}{-1/x^2} = -x, $$

$$ \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} -x $$

So as

$$ \lim_{x \to 0^+} -x = 0, $$

$$ \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = 0$$

Final Step

Recall that

$$ \lim_{x \to 0^+} \frac{\ln(x)}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} $$

and

$$x \ln(x) = \frac{\ln(x)}{1/x} $$

Since

$$ \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = 0,$$

$$ \lim_{x \to 0^+} \frac{\ln(x)}{1/x} = 0 $$

So as

$$x \ln(x) = \frac{\ln(x)}{1/x}, $$

$$ \lim_{x \to 0^+} x \ln(x) = \lim_{x \to 0^+} \frac{\ln(x)}{1/x} $$

$$ = 0 $$

That is the answer to the problem!