AP Calculus AB and BC

Limit of x ln x as x Approaches 0

The limit of x ln x as x approaches 0 from the right is 0. Direct substitution gives 0 times negative infinity, an indeterminate product, so rewrite it as ln x over 1/x. That is a negative infinity over infinity quotient, and one round of L'Hopital collapses it to negative x, which goes to 0.

lim⁡x→0+xln⁡x=0\lim_{x \to 0^+} x\ln x = 0

Settled by rewriting as a quotient then L'Hopital.

Turning the product into a quotient

L'Hopital's rule is stated for quotients, so the product has to be rearranged. Send the polynomial factor downstairs as its reciprocal, which leaves the logarithm on top.

lim⁡x→0+xln⁡x=lim⁡x→0+ln⁡x1x\lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \frac{\ln x}{\frac{1}{x}}

The numerator now runs to −∞-\infty and the denominator to ∞\infty, a genuine ∞∞\frac{\infty}{\infty} form, so differentiate top and bottom separately.

ddx[ln⁡x]ddx[1x]=1x−1x2=−x\frac{\frac{d}{dx}\left[\ln x\right]}{\frac{d}{dx}\left[\frac{1}{x}\right]} = \frac{\frac{1}{x}}{-\frac{1}{x^2}} = -x
lim⁡x→0+xln⁡x=lim⁡x→0+(−x)=0\lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} (-x) = 0

Pick the rewrite that simplifies

The other rearrangement, x1ln⁡x\frac{x}{\frac{1}{\ln x}}, is a legal 00\frac{0}{0} form, but L'Hopital turns it into −x(ln⁡x)2-x(\ln x)^2, which is worse than what you started with. Put the logarithm in the numerator so that differentiating removes it.

Why substitution fails

Substituting x=0x = 0 gives 0⋅(−∞)0 \cdot (-\infty). The first factor is crushing the product toward 0 while the second is stretching it toward −∞-\infty, and the form records nothing about which effect is stronger. It states a race, not a value.

The polynomial wins. xx reaches 0 faster than ln⁡x\ln x falls, because the logarithm is the slowest mover in the standard library: at x=10−6x = 10^{-6} the factor xx has shrunk by six orders of magnitude while ln⁡x\ln x has only reached about −13.8-13.8, so the product sits near −1.4⋅10−5-1.4 \cdot 10^{-5}.

The limit law that a product of limits equals the product of the limits does not rescue you either, since it requires both limits to be finite numbers. That is what pushes the problem onto L'Hopital's rule.

The mistake students make

  • Answering −∞-\infty on the grounds that ln⁡x\ln x blows up. Size is not the tiebreaker, rate is, and any positive power of xx beats a logarithm at 0.
  • Answering 0 by declaring that 0 times anything is 0. True for finite factors, false as a limit statement: x⋅1x2x \cdot \frac{1}{x^2} also looks like 0⋅∞0 \cdot \infty and runs to ∞\infty.
  • Differentiating the product with the product rule. L'Hopital differentiates the numerator and the denominator of a quotient separately, and never the expression as a whole.
  • Losing the sign on ddx[1x]=−1x2\frac{d}{dx}\left[\frac{1}{x}\right] = -\frac{1}{x^2}. Dropping that minus turns the simplified quotient into xx instead of −x-x, which still gives 0 here but will not on the next problem.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

Why can I not apply L'Hopital to x ln x directly?

The rule is stated for f(x)g(x)\frac{f(x)}{g(x)} in 00\frac{0}{0} or ∞∞\frac{\infty}{\infty} form, and a product is neither. Moving one factor into the denominator as its reciprocal changes nothing about the value and makes the rule apply.

Does the same answer hold for x^2 ln x?

Yes, it is also 0. Rewrite as ln⁡x1x2\frac{\ln x}{\frac{1}{x^2}} and L'Hopital gives −x22-\frac{x^2}{2}, which goes to 0. Any positive power of xx overwhelms a logarithm as x→0+x \to 0^+.

Is there a left-hand limit?

No. ln⁡x\ln x requires x>0x > 0, so the function has no values to the left of 0 and only the right-hand limit is defined.