L'Hôpital's Rule

In a 0/00/0 or ∞/∞\infty/\infty form, differentiating the numerator and the denominator keeps the limit. Drive xx toward the limit point and watch f(x)g(x)\frac{f(x)}{g(x)} track f′(x)g′(x)\frac{f'(x)}{g'(x)}.

lim⁡x→af(x)g(x)=lim⁡x→af′(x)g′(x)\lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)}
lim⁡x→0sin⁡xx=lim⁡x→0cos⁡x=1,lim⁡x→∞ln⁡xx=lim⁡x→∞1x=0\lim_{x \to 0}\frac{\sin x}{x} = \lim_{x \to 0}\cos x = 1, \qquad \lim_{x \to \infty}\frac{\ln x}{x} = \lim_{x \to \infty}\frac{1}{x} = 0

Part of these learning paths

Examples & Exercises

Prove lim⁡x→0sin⁡xx=1\lim_{x \to 0}\frac{\sin x}{x} = 1 by L'Hôpital's rule. (State the 0/00/0 hypotheses.)

lim⁡x→0sin⁡xx=lim⁡x→0(sin⁡x)′(x)′=lim⁡x→0cos⁡x=cos⁡0=1\lim_{x \to 0}\frac{\sin x}{x} = \lim_{x \to 0}\frac{(\sin x)'}{(x)'} = \lim_{x \to 0}\cos x = \cos 0 = 1
x≈0.0238x \approx 0.0238 (a 0/00/0 form): sin⁡xx≈0.9999\frac{\sin x}{x} \approx 0.9999, cos⁡x1≈0.9997\frac{\cos x}{1} \approx 0.9997 — dragging right, both stick to 1.
Show proof
  1. Put f(x)=sin⁡xf(x) = \sin x and g(x)=xg(x) = x: then f(0)=g(0)=0f(0) = g(0) = 0 and g′(x)=1≠0g'(x) = 1 \neq 0 — a 0/00/0 form. For x>0x > 0, Cauchy's theorem on [0,x][0, x] gives sin⁡x−sin⁡0x−0=cos⁡ξx1\frac{\sin x - \sin 0}{x - 0} = \frac{\cos \xi_x}{1} for some ξx∈(0,x)\xi_x \in (0, x).
  2. Since 0<ξx<x0 < \xi_x < x, squeezing gives ξx→0\xi_x \to 0 as x→0+x \to 0^+. Cosine is continuous (differentiable implies continuous), so cos⁡ξx→cos⁡0=1\cos \xi_x \to \cos 0 = 1. That is L'Hôpital's rule for the 0/00/0 case: the limit of the differentiated ratio equals the limit of the original ratio.
  3. For x<0x < 0, the same argument on [x,0][x, 0] yields ξx∈(x,0)\xi_x \in (x, 0) and likewise cos⁡ξx→1\cos \xi_x \to 1. Hence lim⁡x→0sin⁡xx=1\lim_{x \to 0}\frac{\sin x}{x} = 1.

Prove lim⁡x→∞ln⁡xx=0\lim_{x \to \infty}\frac{\ln x}{x} = 0. (An ∞/∞\infty/\infty form: check the g→∞g \to \infty and g′≠0g' \neq 0 hypotheses.)

f(x)g(x)=f′(ξx)g′(ξx)(1−g(A)g(x))+f(A)g(x)→x→∞L\frac{f(x)}{g(x)} = \frac{f'(\xi_x)}{g'(\xi_x)}\left(1 - \frac{g(A)}{g(x)}\right) + \frac{f(A)}{g(x)} \xrightarrow[x\to\infty]{} L
(ln⁡x)′(x)′=1/x1=1x→0⇒lim⁡x→∞ln⁡xx=0\frac{(\ln x)'}{(x)'} = \frac{1/x}{1} = \frac{1}{x} \to 0 \quad\Rightarrow\quad \lim_{x\to\infty}\frac{\ln x}{x} = 0
x≈158.5x \approx 158.5 (an ∞/∞\infty/\infty form): ln⁡xx≈0.0320\frac{\ln x}{x} \approx 0.0320, 1x≈0.00631\frac{1}{x} \approx 0.00631 — dragging right, both stick to 0.
Show proof
  1. First the ∞/∞\infty/\infty lemma: suppose f,gf, g are differentiable on (A,∞)(A, \infty) with g′≠0g' \neq 0, g(x)→∞g(x) \to \infty, and f′(x)g′(x)→L\frac{f'(x)}{g'(x)} \to L (finite). For x>Ax > A, Cauchy's theorem on [A,x][A, x] gives f(x)−f(A)g(x)−g(A)=f′(ξx)g′(ξx)\frac{f(x) - f(A)}{g(x) - g(A)} = \frac{f'(\xi_x)}{g'(\xi_x)} for some ξx∈(A,x)\xi_x \in (A, x).
  2. Rearranging, f(x)g(x)=f′(ξx)g′(ξx)(1−g(A)g(x))+f(A)g(x)\frac{f(x)}{g(x)} = \frac{f'(\xi_x)}{g'(\xi_x)}\left(1 - \frac{g(A)}{g(x)}\right) + \frac{f(A)}{g(x)}. Given ε>0\varepsilon > 0, choose AA so that ∣f′(t)g′(t)−L∣<ε2\left|\frac{f'(t)}{g'(t)} - L\right| < \frac{\varepsilon}{2} for t>At > A; then ξx>A\xi_x > A gives ∣f′(ξx)g′(ξx)−L∣<ε2\left|\frac{f'(\xi_x)}{g'(\xi_x)} - L\right| < \frac{\varepsilon}{2}. Since g(x)→∞g(x) \to \infty, for large xx we have ∣g(A)g(x)∣<1\left|\frac{g(A)}{g(x)}\right| < 1 and ∣f(A)−Lg(A)∣∣g(x)∣\frac{|f(A) - L g(A)|}{|g(x)|} small, so ∣f(x)g(x)−L∣<ε\left|\frac{f(x)}{g(x)} - L\right| < \varepsilon. With ε\varepsilon arbitrary, f(x)g(x)→L\frac{f(x)}{g(x)} \to L — L'Hôpital's rule for the ∞/∞\infty/\infty case.
  3. Apply this to f(x)=ln⁡xf(x) = \ln x and g(x)=xg(x) = x: g′(x)=1≠0g'(x) = 1 \neq 0, g(x)→∞g(x) \to \infty, and f′(x)g′(x)=1/x1=1x→0\frac{f'(x)}{g'(x)} = \frac{1/x}{1} = \frac{1}{x} \to 0. Hence lim⁡x→∞ln⁡xx=0\lim_{x \to \infty}\frac{\ln x}{x} = 0.

Prove lim⁡x→0+xln⁡x=0\lim_{x \to 0^+} x \ln x = 0. (Hint: rewrite it as ln⁡x1/x\frac{\ln x}{1/x}.)

xln⁡x=ln⁡x1/x,(ln⁡x)′(1/x)′=1/x−1/x2=−x→x→0+0x \ln x = \frac{\ln x}{1/x}, \qquad \frac{(\ln x)'}{(1/x)'} = \frac{1/x}{-1/x^2} = -x \xrightarrow[x\to 0^+]{} 0
The same ∞/∞\infty/\infty machinery — at x=1000x = 1000, ln⁡xx≈0.00691\frac{\ln x}{x} \approx 0.00691, 1x=0.001\frac{1}{x} = 0.001. Rewriting xln⁡x=ln⁡x1/xx\ln x = \frac{\ln x}{1/x} gives f′g′=−x→0\frac{f'}{g'} = -x \to 0 (e.g. at x=0.01x = 0.01, xln⁡x≈−0.0461x\ln x \approx -0.0461, −x=−0.01-x = -0.01).
Show proof
  1. For x>0x > 0, xln⁡x=ln⁡x1/xx \ln x = \frac{\ln x}{1/x}. Put f(x)=ln⁡xf(x) = \ln x and g(x)=1/xg(x) = 1/x: then f(x)→−∞f(x) \to -\infty and g(x)→+∞g(x) \to +\infty, an ∞/∞\infty/\infty form, and g′(x)=−1x2≠0g'(x) = -\frac{1}{x^2} \neq 0.
  2. f′(x)g′(x)=1/x−1/x2=−x→x→0+0\frac{f'(x)}{g'(x)} = \frac{1/x}{-1/x^2} = -x \xrightarrow[x \to 0^+]{} 0. Applying problem 2's lemma adapted to x→0+x \to 0^+ (the same argument on [x,A][x, A] with 0<x<A0 < x < A) gives lim⁡x→0+ln⁡x1/x=0\lim_{x \to 0^+}\frac{\ln x}{1/x} = 0.
  3. Hence lim⁡x→0+xln⁡x=0\lim_{x \to 0^+} x \ln x = 0. The numbers agree: at x=0.01x = 0.01, xln⁡x=0.01⋅ln⁡0.01≈−0.0461x\ln x = 0.01 \cdot \ln 0.01 \approx -0.0461 and the derivative ratio is −x=−0.01-x = -0.01, both tending to 0 as x→0+x \to 0^+.

9. Differentiation