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Wize University Calculus 1 Textbook > Applications of Differentiation

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]




Extra Practice
L'Hopital's Rule: General Indeterminate Forms
Evaluate the limit L=lim⁡x→0+xsin⁡ xL=\lim\limits_{x\rightarrow0^+}x^{\sin\ x}.
L'Hospital's Rule: Indeterminate Forms
Evaluate lim⁡ x→∞(1−kx)3x\lim\ _{x\rightarrow\infty}\left(1-\frac{k}{x}\right)^{3x}.
Indeterminate Forms
Evaluate the limit if it exists.
lim⁡x→∞ (1+1x)x\lim_{x\to\infty}\ \left(1+\frac{1}{x}\right)^x.
L'Hospital's Rule: Indeterminate Forms
Evaluate lim⁡ x→∞(1−kx)3x\lim\ _{x\rightarrow\infty}\left(1-\frac{k}{x}\right)^{3x}.
L'Hospital's Rule: Indeterminate Forms
Find lim⁡x→0+(ln⁡x)(sin⁡x)\lim_{x\rightarrow0^+}\left(\ln x\right)\left(\sin x\right).
L'Hospital's Rule: Indeterminate Forms
Evaluate lim⁡x→0+(e1x)1ln⁡x\lim_{x\rightarrow0^+}\left(e^{\frac{1}{x}}\right)^{^{\frac{1}{\ln x}}}.
Indeterminate Forms
Evaluate the limit: lim⁡x→0+(2x)x2\displaystyle \lim_{x\rightarrow 0^+}(2x)^{\frac{x}{2}}
Indeterminate Forms: Polynomial and Ln
Evaluate the limit: lim⁡x→0+xaln⁡x\displaystyle \lim_{x\rightarrow 0^+}x^a\ln{x}, where a>0a>0.
Practice: Fractions with Trig
Q.\textbf{Q.} Evaluate: lim⁡x→0+(2x−2sin⁡x)\displaystyle \lim_{x\rightarrow 0^+}\left(\frac{2}{x}-\frac{2}{\sin{x}}\right)
Evaluate the following limit: lim⁡x→0+(2x)x2\displaystyle \lim_{x\rightarrow 0^+}(2x)^{\frac{x}{2}}
Evaluate the following limit: lim⁡x→0+x32ln⁡x\displaystyle \lim_{x\rightarrow0^+}x^{\frac{3}{2}}\ln{x}
Practice: $f^g$ Form with Trig
Q.\textbf{Q.} Evaluate the following limit: lim⁡x→0+cos⁡xcot⁡x\displaystyle \lim_{x\rightarrow 0^+} \cos{x}^{\cot{x}}
Indeterminate Forms
Evaluate the limit: lim⁡x→∞[1+sin⁡(2x)]2x\displaystyle \lim_{x\rightarrow\infty}\left[1+\sin\left(\frac{2}{x}\right)\right]^{2x}
Practice: L'Hospital's Rule
Evaluate lim⁡x→0+(e1x)1ln⁡x\lim_{x\rightarrow0^+}\left(e^{\frac{1}{x}}\right)^{^{\frac{1}{\ln x}}}.
Practice: L'Hospital's Rule
Evaluate lim⁡x→∞ x3 sin⁡(1x2)\displaystyle \lim_{x\to\infty}\ x^3\ \sin\left(\frac{1}{x^2}\right).
Practice: L'Hospital's Rule
Evaluate lim⁡x→∞ x3 sin⁡(1x2)\displaystyle \lim_{x\to\infty}\ x^3\ \sin\left(\frac{1}{x^2}\right).
General Indeterminate Forms
Evaluate:
lim⁡x→+∞2x+19x2−2x−6\lim_{x \rightarrow +\infty} \frac{2x + 1}{\sqrt{9x^2-2x - 6}}
Practice: L'Hospital's Rule
Evaluate lim⁡x→0+(e1x)1ln⁡x\lim_{x\rightarrow0^+}\left(e^{\frac{1}{x}}\right)^{^{\frac{1}{\ln x}}}.
Practice: L'Hospital's Rule
Evaluate lim⁡x→∞ x3 sin⁡(1x2)\displaystyle \lim_{x\to\infty}\ x^3\ \sin\left(\frac{1}{x^2}\right).
Evaluate lim⁡x→0+2ln⁡xln⁡(ex−1)\lim_{x\rightarrow0^+}\frac{2\ln x}{\ln\left(e^x-1\right)}
General Indeterminate Forms
Evaluate
lim⁡x→0+xaln⁡x where a>0\lim\limits_{x\to0^+} x^a\ln x \text{ where } a>0
l'Hopital's Rule: General Indeterminate Forms
Evaluate
lim⁡x→0+x3ln⁡x\lim\limits_{x\to0^+}x^3\ln x
Evaluate the following limit: lim⁡x→0+(2x)x2\displaystyle \lim_{x\rightarrow 0^+}(2x)^{\frac{x}{2}}
Evaluate the following limit: lim⁡x→0+x32ln⁡x\displaystyle \lim_{x\rightarrow0^+}x^{\frac{3}{2}}\ln{x}
Evaluate the following limit: lim⁡x→0+x32ln⁡x\displaystyle \lim_{x\rightarrow0^+}x^{\frac{3}{2}}\ln{x}
Evaluate the following limit: lim⁡x→0+(2x)x2\displaystyle \lim_{x\rightarrow 0^+}(2x)^{\frac{x}{2}}
Indeterminate Forms: Power and Ln
Evaluatelim⁡x→0+x32ln⁡x\displaystyle \lim_{x\rightarrow 0^+} x^{\frac{3}{2}}\ln{x}
Practice: $f^g$ Form with Trig
Q.\textbf{Q.} Evaluate the following limit: lim⁡x→0+cos⁡xcot⁡x\displaystyle \lim_{x\rightarrow 0^+} \cos{x}^{\cot{x}}
Practice: Fractions with Trig
Q.\textbf{Q.} Evaluate: lim⁡x→0+(2x−2sin⁡x)\displaystyle \lim_{x\rightarrow 0^+}\left(\frac{2}{x}-\frac{2}{\sin{x}}\right)
Indeterminate Forms
Evaluate the limit: lim⁡x→∞[1+sin⁡(2x)]2x\displaystyle \lim_{x\rightarrow\infty}\left[1+\sin\left(\frac{2}{x}\right)\right]^{2x}
Indeterminate Forms: Polynomial and Ln
Evaluate the limit: lim⁡x→0+xaln⁡x\displaystyle \lim_{x\rightarrow 0^+}x^a\ln{x}, where a>0a>0.
Indeterminate Forms
Evaluate the limit: lim⁡x→0+(2x)x2\displaystyle \lim_{x\rightarrow 0^+}(2x)^{\frac{x}{2}}
Determine lim⁡x→∞(1+7x)x5\lim_{x\rightarrow\infty}\left(1+\frac{7}{x}\right)^{\frac{x}{5}}.
Determine lim⁡ x→∞(x)1x2\lim\ _{x\rightarrow\infty}\left(\sqrt{x}\right)^{\frac{1}{x^2}}.
Evaluate the limit lim⁡x→∞(1+sin⁡2x)2x\displaystyle \lim_{x\rightarrow \infty}\left(1+\sin{\frac{2}{x}}\right)^{2x}.
Evaluate the limit: lim⁡x→0+(2x−2sin⁡x)\displaystyle \lim_{x\rightarrow 0^+}\left(\frac{2}{x}-\frac{2}{\sin{x}}\right)
Solve the limit lim⁡x →0(1−2x)1x2\displaystyle\lim_{x\ \to0}\left(1-2x\right)^{\frac{1}{x^2}}.
Find lim⁡x→0+(ln⁡x)(sin⁡x)\lim_{x\rightarrow0^+}\left(\ln x\right)\left(\sin x\right).
🦊 TRICKY! Determine lim⁡x→2+(x−2)1x−2\lim_{x\rightarrow2^+}\left(x-2\right)^{\frac{1}{x-2}}.
L'Hopital's Rule: General Indeterminate Forms
Evaluate the limit L=lim⁡x→0+xsin⁡ xL=\lim\limits_{x\rightarrow0^+}x^{\sin\ x}.
l'Hopital's Rule: General Indeterminate Forms
Evaluate
lim⁡x→∞(1+7x)x/5\lim_{x\to \infty}\left(1+\frac{7}{x}\right)^{x/5}
Evaluate lim⁡x→0+x32ln⁡x\lim_{x\rightarrow0^+}x^{\frac{3}{2}}\ln x
Determine lim⁡x→∞(1+7x)x5\lim_{x\rightarrow\infty}\left(1+\frac{7}{x}\right)^{\frac{x}{5}}
Indeterminate Forms
Evaluate the limit if it exists.
lim⁡x→∞ (1+1x)x\lim_{x\to\infty}\ \left(1+\frac{1}{x}\right)^x.
Indeterminate Forms
Evaluate the limit, if it exists.
lim⁡x→3 1x−3−e1x−3\lim_{x\to3}\ \frac{1}{x-3}-e^{\frac{1}{x-3}}.
Indeterminate Forms
Evaluate the limit, if it exists.
lim⁡x→−2 (x+2)ln⁡∣x+2∣\lim_{x\to-2}\ \left(x+2\right)\ln\left|x+2\right|
L'Hospital's Rule: Indeterminate Forms
Evaluate lim⁡ x→∞(1−kx)3x\lim\ _{x\rightarrow\infty}\left(1-\frac{k}{x}\right)^{3x}.
L'Hospital's Rule: Indeterminate Forms
Evaluate lim⁡x→0+(e1x)1ln⁡x\lim_{x\rightarrow0^+}\left(e^{\frac{1}{x}}\right)^{^{\frac{1}{\ln x}}}.
L'Hospital's Rule: Indeterminate Forms
Find lim⁡x→0+(ln⁡x)(sin⁡x)\lim_{x\rightarrow0^+}\left(\ln x\right)\left(\sin x\right).
Evaluate lim⁡x→0+x32ln⁡x\lim_{x\rightarrow0^+}x^{\frac{3}{2}}\ln x
Practice: L'hopital's rule
Q.\textbf{Q.} Evaluate the following limits.
1. lim⁡x→0+cos⁡xcot⁡x\displaystyle \lim_{x\rightarrow 0^+} \cos{x}^{\cot{x}}
2. lim⁡x→∞[x2+1−x]\displaystyle \lim_{x\rightarrow \infty}\left[\sqrt{x^2+1}-x\right]