In one line
Convert them to \(\frac00\) or \(\frac\infty\infty\): rewrite products and differences as a single fraction, and for powers take logarithms first; then apply L'Hospital's rule.
Explained simply
L'Hospital's rule only understands fractions. The other indeterminate forms are fractions in disguise. A product can be written as a quotient by moving one factor to the bottom as its reciprocal. A difference becomes a fraction by taking a common denominator. A power \(y=f^g\) becomes a product after taking log: \(\log y=g\log f\).
Technically: \(0\cdot\infty\): write \(fg=\dfrac{f}{1/g}\). \(\infty-\infty\): combine into one fraction. \(1^\infty,0^0,\infty^0\): let \(y=f^g\), find \(L=\lim g\log f\) (now \(0\cdot\infty\)), then the answer is \(e^L\). This is a self-learning topic.
Why can't we just "plug in"? Because \(1^\infty\), for instance, hides a tug of war: the base is pulling the value towards 1, the exponent is pulling it towards infinity or zero. Depending on the exact speeds, the result can be any positive number, and the famous limit \((1+x)^{1/x}\to e\) is one such case. That is why these forms must be worked out, never guessed.
How it works
- Identify the form by substitution.
- Rewrite as \(\frac00\) or \(\frac\infty\infty\) using the conversions above.
- Apply L'Hospital's rule as many times as needed.
- For power forms, remember to convert back: limit of \(y\) is \(e^{L}\).
Key formula
For powers: \(\lim f^g=e^{\lim g\log f}\). Standard results: \(\displaystyle\lim_{x\to0}(1+x)^{1/x}=e\), \(\displaystyle\lim_{x\to0^+}x^x=1\).
Quick example
- \(0\cdot\infty\): \(\displaystyle\lim_{x\to0^+}x\log x=\lim\frac{\log x}{1/x}=\lim\frac{1/x}{-1/x^2}=\lim(-x)=0\).
- \(0^0\): \(y=x^x\), \(\log y=x\log x\to0\), so \(\lim x^x=e^0=1\).
- \(1^\infty\): \(y=(1+x)^{1/x}\), \(\log y=\dfrac{\log(1+x)}x\to\dfrac{1/(1+x)}{1}\to1\), so the limit is \(e\).
- \(\infty-\infty\): \(\displaystyle\lim_{x\to0}\left(\frac1x-\frac1{\sin x}\right)=\lim\frac{\sin x-x}{x\sin x}\to\lim\frac{\cos x-1}{\sin x+x\cos x}\to\lim\frac{-\sin x}{2\cos x-x\sin x}=0\).
Common mistakes
- Forgetting to exponentiate at the end of a power form, giving \(L\) instead of \(e^L\).
- Treating \(1^\infty\) as 1; it is indeterminate.
- Applying L'Hospital's rule directly to a product or difference.
How MU asks it
As a self-learning topic expect at most a 3–5-mark part: "Evaluate \(\displaystyle\lim_{x\to0}(\cos x)^{1/x^2}\)" (answer \(e^{-1/2}\)) or "\(\displaystyle\lim_{x\to\pi/2}(\sec x-\tan x)\)" (answer 0). Name the form, show the conversion step, apply the rule, and convert back for powers.