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Wize AP Calculus (BC) Textbook > Applications of Differentiation: Contextual

Solving General Indeterminate Forms

L’Hopital’s Rule & Indeterminate FormsPrevious Section
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Solving General Indeterminate Forms

Using only L'Hoptial's Rule is not enough to solve every type of limit. Depending on the situation, we may have to perform some algebra to effectively apply L'Hoptial's Rule.

Techniques to Solve General Indeterminate Forms

  1. Limits of the form 00\displaystyle \frac{0}{0} and ±∞±∞\displaystyle \frac{\pm\infty}{\pm\infty} can be solved using L'Hopital's Rule, factoring the highest power at the numerator and denominator or cancelling common factors.
  2. Limits of the form 0⋅∞0\cdot\infty can be written as f(x)g(x)=g(x)1f(x)=f(x)1g(x)\displaystyle f(x)g(x)=\frac{g(x)}{\frac{1}{f(x)}}=\frac{f(x)}{\frac{1}{g(x)}} and going back to Case 1.
  3. Limits of the form 00,∞00^0,\infty^0 or 1∞1^\infty can be reduced to Case 2 by taking the natural logarithms on both sides.
  4. Limits of the form ∞−∞\infty-\infty can be solved by either putting on the same denominator, by factoring or by multiplying by the conjugate, and then going back to Case 1.
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Example: Solving General Indeterminate Forms

Find lim⁡x→0+xaln⁡x\displaystyle \lim_{x\rightarrow 0^+}x^a\ln{x}, where a>0a>0

lim⁡x→0+xaln⁡x=0×−∞ (indeterminate form)\displaystyle \lim_{x\rightarrow 0^+}x^a\ln{x}=0\times -\infty\text{ (indeterminate form)}

Rewrite as a fraction to use L'Hopital's Rule.

lim⁡x→0+xaln⁡x=lim⁡x→0+ln⁡xx−a=−∞∞ (indeterminante form)\displaystyle \lim_{x\rightarrow 0^+}x^a\ln{x}=\displaystyle \lim_{x\rightarrow 0^+}\frac{\ln{x}}{x^{-a}}=\frac{-\infty}{\infty}\text{ (indeterminante form)}

We can use L'Hopital's Rule:

lim⁡x→0+ln⁡xx−a=L′Hopital′slim⁡x→0+1x−ax−a−1=−1alim⁡x→0+xa=0\begin{array}{cl} &\displaystyle \lim_{x\rightarrow 0^+}\frac{\ln{x}}{x^{-a}}\\\\ \overset{L'Hopital's}{=}&\displaystyle \lim_{x\rightarrow 0^+}\frac{\frac{1}{x}}{-ax^{-a-1}}\\\\ =&\displaystyle \frac{-1}{a}\lim_{x\rightarrow 0^+}x^a\\\\ =&\boxed0 \end{array}

(since a>0a>0)

Evaluatelim⁡x→0+cos⁡xcot⁡x\displaystyle \lim_{x\rightarrow 0^+} \cos{x}^{\cot{x}}.



Evaluate lim⁡x→∞[x2+1−x]\displaystyle \lim_{x\rightarrow \infty}\left[\sqrt{x^2+1}-x\right]