Hôpital’s Theorem

Let f, g be two functions defined in a neighborhood of c, except possibly at c, such that

\[\lim_{x \to c} f(x) = \lim_{x \to c} g(x) = L,\]

with L = 0 or L = +∞ or L = −∞.
If f and g are differentiable in a neighborhood of c, except possibly at c, with g′ ≠ 0, and if the following limit exists (finite or infinite):

\[\lim_{x \to c} \frac{f'(x)}{g'(x)},\]

then also

\[\lim_{x \to c} \frac{f(x)}{g(x)}\]

exists and equals the previous one.

Exercises

Ex 1.

\[\lim_{x \to 0}\frac{e^{x^{x}}-e}{x}.\]

Solution
Indeterminate form 0/0. Apply Hôpital:

\[(x^{x})' = (e^{x \log x})' = (\log x +1)x^{x}.\]

So

\[\lim_{x \to 0}\frac{e^{x^{x}}-e}{x} \overset{H}{=} \lim_{x \to 0} (\log x+1)x^{x} e^{x^x} = -\infty .\]

Indeed,

\[\lim_{x \to 0} x^x = e^0 = 1.\]

Ex 2.

\[\lim_{x \to 0}\Bigl(\frac{\sin x}{x}\Bigr)^{1/x^{2}}.\]

Solution

\[\lim_{x \to 0}\Bigl(\frac{\sin x}{x}\Bigr)^{1/x^{2}} = \lim_{x \to 0} e^{\frac{\log(\sin x/x)}{x^2}} = e^{-1/6}.\]

Ex 3.

\[\lim_{x \to 0}\frac{e^{\sin x}-\cos x}{e^{\cos x}-e\log(x+e)} .\]

Solution
Indeterminate form 0/0:

\[\lim_{x \to 0}\frac{e^{\sin x}-\cos x}{e^{\cos x}-e\log(x+e)} \overset{H}{=} \lim_{x \to 0} \frac{\cos x \, e^{\sin x} + \sin x}{-\sin x \, e^{\cos x} - \tfrac{e}{x+e}} = -1 .\]

Ex 4.

\[\lim_{x \to -1}\frac{e^{\sqrt{1-3x}}-e^{\sqrt{3-x}}}{e^{x}-e^{2x+1}} .\]

Solution

\[\lim_{x \to -1}\frac{e^{\sqrt{1-3x}}-e^{\sqrt{3-x}}}{e^{x}-e^{2x+1}} \overset{H}{=} \frac{-\tfrac{3}{4}e^{2}+\tfrac{e^{2}}{4}}{\tfrac{1}{e}-\tfrac{2}{e}} = \frac{e^3}{2}.\]

Ex 5.

\[\lim_{x \to 1^{-}} \log(x)\log(1-x).\]

Solution

\[\lim_{x \to 1^{-}} \log(x)\log(1-x) = \lim_{x \to 1^{-}} \frac{\log(1-x)}{1/\log x} \overset{H}{=} \lim_{x \to 1^{-}} \frac{-\tfrac{1}{1-x}}{-1/(x \log^2 x)} = 0.\]

Ex 6.

\[\lim_{x \to 3^{+}}\frac{\sqrt{x}-\sqrt{3}+\sqrt{x-3}}{\sqrt{x^{2}-9}} .\]

Solution

\[\lim_{x \to 3^{+}}\frac{\sqrt{x}-\sqrt{3}+\sqrt{x-3}}{\sqrt{x^{2}-9}} = \frac{1}{\sqrt{6}} .\]

Indeed,

\[\lim_{x \to 3^+} \frac{\sqrt{x}-\sqrt{3}}{\sqrt{x-3}} \overset{H}{=} \lim_{x \to 3^+} \frac{1/(2\sqrt{x})}{1/(2\sqrt{x-3})} = 0.\]

Ex 7.

\[\lim_{x \to 0^{+}}\frac{x^{2}-\arctan(x^{2})}{x(1-\cos x)^{3}} .\]

Solution

\[\lim_{x \to 0^{+}}\frac{x^{2}-\arctan(x^{2})}{x(1-\cos x)^{3}} \overset{H}{=} \lim_{x \to 0^+} \frac{2x - \tfrac{2x}{1+x^4}}{(1-\cos x)^3+3x(1-\cos x)^2\sin x} = +\infty .\]

Ex 8.

\[\lim_{x \to 0}\frac{e^{x}-\cos x-\sin x}{e^{x^{2}}-e^{x^{3}}} .\]

Solution
Taylor near 0:

\[e^x = 1+x+\tfrac{x^2}{2}+\tfrac{x^3}{6}+o(x^3), \quad \cos x = 1-\tfrac{x^2}{2}+o(x^2), \quad \sin x = x-\tfrac{x^3}{6}+o(x^3).\]

So numerator ∼ x² + x³/6, denominator ∼ x² − x³.
Hence limit = 1.

Ex 9.

\[\lim_{x \to 0}\frac{1-\cos x+\log(\cos x)}{x^{4}} .\]

Solution
Expansions:

\[\cos x = 1-\tfrac{x^2}{2}+\tfrac{x^4}{24}+o(x^4),\] \[\log(\cos x) = -\tfrac{x^2}{2} - \tfrac{x^4}{12}+o(x^4).\]

So numerator ∼ −⅛x⁴.

\[\lim_{x \to 0}\frac{1-\cos x+\log(\cos x)}{x^4} = -\tfrac{1}{8}.\]

Ex 10.

\[\lim_{x \to 0}\frac{e^{x^{2}}-\cos^{2}x}{\sin^{2}x}.\]

Solution (two methods)

By Hôpital:

\[\lim_{x \to 0}\frac{e^{x^{2}}-\cos^{2}x}{\sin^{2}x} \overset{H}{=} \lim_{x \to 0}\frac{2xe^{x^2}+2\cos x \sin x}{2\cos x \sin x} = \lim_{x \to 0} \frac{x}{\sin x}\cdot\frac{e^{x^2}}{\cos x} +1 = 2.\]

By Taylor:

\[e^{x^2}=1+x^2+o(x^2), \quad \cos^2 x = 1-x^2+o(x^2).\]

So numerator ∼ 2x², denominator ∼ x², ratio = 2.

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