Indeterminate Forms and LHospitals Rule These expressions are not real numbers. However, there are many more indeterminate forms out Suppose \(f\) and \(g\) are differentiable functions over an open interval containing \(a\), except possibly at \(a\). A slight change of the previous example shows that it is possible that, \begin{align*} \lim\limits_{x\rightarrow a}f(x) &=0 & \text{and}&& \lim\limits_{x\rightarrow a}g(x) &= 0 \end{align*}, \begin{align*} \lim\limits_{x\rightarrow a}\frac{f(x)}{g(x)} && \text{or}&& \lim\limits_{x\rightarrow a}\frac{f'(x)}{g'(x)} \end{align*}, \begin{gather*} a=0\qquad \qquad f(x)=x\sin\frac{1}{x} \qquad g(x)= x \end{gather*}, Then (with a quick application of the squeeze theorem), \begin{align*} \lim_{x\rightarrow 0}\frac{f(x)}{g(x)} &=\lim_{x\rightarrow 0}\frac{x\sin\frac{1}{x} }{x} =\lim_{x\rightarrow 0} \sin\frac{1}{x} \end{align*}, \begin{align*} \lim_{x\rightarrow 0}\frac{f'(x)}{g'(x)} &= \lim_{x\to 0} \frac{\sin\frac{1}{x} - \frac{1}{x}\cos\frac{1}{x} }{x^2} \end{align*}. Rather, these expressions represent forms that arise when finding limits. Indeterminate Forms and L'Hospital's Rule Section 4.10 : L'Hospital's Rule and Indeterminate Forms a. We proceed as follows. }\) To do this, we make use of the Generalised Mean-Value Theorem (Theorem 3.4.38) that was used to prove Equation 3.4.33. &=\dfrac{f(a)}{g(a)} & & \text{By the definition of the derivative} \\[4pt] Since the least common denominator is \(x^2\tan x,\) we have. \nonumber \], Evaluate \[\lim_{x0}\dfrac{x}{\tan x}. \nonumber \], Lets consider the first option. \nonumber \], \(\dfrac{d}{dx}\big(\ln x\big)=\dfrac{1}{x}\). \end{align*}, \begin{align*} \lim_{x\to 0} \frac{f(x)}{g(x)} &= \lim_{x\to 0} \frac{f'(x)}{g'(x)} \end{align*}, \begin{gather*} \lim_{x \to 0} \frac{2x \cos(x^2)}{\sin(x)} \end{gather*}, \begin{align*} h(x) &= 2x\cos(x^2) & h'(x) &= 2\cos(x^2) - 4x^2\sin(x^2) & h'(0) &=2\\ \ell(x) &= \sin(x) & \ell'(x) &= \cos(x) & \ell'(0) &= 1 \end{align*}, \begin{align*} \lim_{x \to 0} \frac{2x \cos(x^2)}{\sin(x)} &= \frac{h'(0)}{\ell'(0)} = 2 \end{align*}, \begin{align*} \lim_{x\to 0} \frac{\sin(x^2)}{1-\cos x} &=\lim_{x \to 0} \frac{2x \cos(x^2)}{\sin(x)} = 2. Therefore, we can apply LHpitals rule. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Accessibility StatementFor more information contact us atinfo@libretexts.org. Suppose that. In your example, the limit is not indeterminate; it's 0. }\), Evaluate \(\lim\limits_{x\rightarrow\infty}x^2e^{-x}\text{. Taking the derivatives of the numerator and the denominator, we have, \[\lim_{x0^+}\dfrac{(\tan x)x^2}{x^2\tan x}=\lim_{x0^+}\dfrac{(\sec^2x)2x}{x^2\sec^2x+2x\tan x}. }\), Evaluate \(\displaystyle\lim_{x \to 0^+} e^{x \log x}\text{. 2 . One application of l'Hpital's rule gives, \begin{align*} \lim_{x\rightarrow 0}\underbrace{\frac{\log\frac{\sin x}{x}}{x^2}}_{ {\mathrm{num}\rightarrow 0}\\{\mathrm{den}\rightarrow 0}} & =\lim_{x\rightarrow 0} \frac {\frac{x}{\sin x}\frac{x\cos x -\sin x}{x^2} }{2x} =\lim_{x\rightarrow 0} \frac{ \frac{x\cos x -\sin x}{x\sin x} }{2x} =\lim_{x\rightarrow 0} \frac{x\cos x -\sin x}{2x^2\sin x} \end{align*}. Consider \(\displaystyle\lim_{x1}\dfrac{x^2+5}{3x+4}.\). WebDenition 1.2. If \(\displaystyle \lim_{xa}f(x)=0\) and \(\displaystyle \lim_{xa}g(x)=0,\) then, \[\lim_{xa}\dfrac{f(x)}{g(x)}=\lim_{xa}\dfrac{f(x)}{g(x)}, \nonumber \], assuming the limit on the right exists or is \(\) or \(\). \nonumber \]. Then, \[\lim_{xa}\dfrac{f(x)}{g(x)}=\lim_{xa}\dfrac{f(x)}{g(x)} \nonumber \]. Example: Find the following limits. Suppose \(\displaystyle\lim_{xa}f(x)=\) (or \(\)) and \(\displaystyle\lim_{xa}g(x)=\) (or \(\)). With this rule, we will be able to evaluate many limits we have not yet been able to determine. Note that the assumption that \(f\) and \(g\) are continuous at \(a\) and \(g(a)0\) can be loosened. Let us return to limits (Chapter 1) and see how we can use derivatives to simplify certain families of limits called indeterminate forms. This then gives, \begin{align*} \lim_{x\rightarrow \frac{\pi}{2}^-}\Big( \underbrace{\sec x}_{\rightarrow +\infty} - \underbrace{\tan x}_{\rightarrow+\infty}\Big) & =\lim_{x\rightarrow \frac{\pi}{2}^-}\underbrace{\frac{1-\sin x} {\cos x}}_{{\mathrm{num}\rightarrow 0} \\{\mathrm{den}\rightarrow 0}} =\lim_{x\rightarrow \frac{\pi}{2}^-}\underbrace{\frac{-\cos x} {-\sin x}}_{{\mathrm{num}\rightarrow 0} \\{\mathrm{den}\rightarrow -1}} =0 \end{align*}. it is not difficult to show that \(e^x\) grows more rapidly than \(x^p\) for any \(p>0\). Calculus I - L'Hospital's Rule and Indeterminate Forms }\) As we take the limit as \(x\to a\text{,}\) we also have that \(c\to a\text{,}\) and so, \begin{align*} \lim_{x\to a^+}\frac{f(x)}{g(x)} &= \lim_{x\to a^+} \frac{f'(c)}{g'(c)} = \lim_{c \to a^+} \frac{f'(c)}{g'(c)} \end{align*}. No, but you don't need to. \(\dfrac{1}{x^2}\dfrac{1}{\tan x}=\dfrac{(\tan x)x^2}{x^2\tan x}\). \begin{gather*} \lim_{x\rightarrow 0}\Big(\frac{\sin x}{x}\Big)^{\frac{1}{x^2}} \end{gather*}, the base, \(\frac{\sin x}{x}\text{,}\) converges to \(1\) (see Example 3.7.3) and the exponent, \(\frac{1}{x^2}\text{,}\) goes to \(\infty\text{. Now rewrite 4, \begin{align*} \frac{f(x)}{g(x)} &= \frac{f(x)}{g(x)} +\underbrace{\left( \frac{f(t)}{g(x)} - \frac{f(t)}{g(x)}\right)}_{=0} + \underbrace{\left(\frac{f(x)-f(t)}{g(x)-g(t)} - \frac{f(x)-f(t)}{g(x)-g(t)}\right)}_{=0}\\ &= \underbrace{\frac{f(x)-f(t)}{g(x)-g(t)}}_\text{ready for GMVT} + \frac{f(t)}{g(x)} + \underbrace{\left( \frac{f(x)}{g(x)} - \frac{f(t)}{g(x)} - \frac{f(x)-f(t)}{g(x)-g(t)} \right)}_\text{we can clean it up}\\ &= \frac{f(x)-f(t)}{g(x)-g(t)} + \frac{f(t)}{g(x)} + \left( \frac{f(x)-f(t)}{g(x)} - \frac{f(x)-f(t)}{g(x)-g(t)} \right)\\ &= \frac{f(x)-f(t)}{g(x)-g(t)} + \frac{f(t)}{g(x)} + \left( \frac{1}{g(x)} - \frac{1}{g(x)-g(t)} \right)\cdot (f(x)-f(t))\\ &= \frac{f(x)-f(t)}{g(x)-g(t)} + \frac{f(t)}{g(x)} + \left( \frac{g(x)-g(t) - g(x)}{g(x)(g(x)-g(t))} \right)\cdot (f(x)-f(t))\\ &= \underbrace{\frac{f(x)-f(t)}{g(x)-g(t)}}_\text{ready for GMVT} + \frac{f(t)}{g(x)} - \frac{g(t)}{g(x)}\cdot \underbrace{\frac{f(x)-f(t)}{g(x)-g(t)}}_\text{ready for GMVT} \end{align*}, Oof! (The indeterminate forms \(0^0\) and \(1^\) can be handled similarly.) WebThen, we first check whether it is an indeterminate form or not by directly putting the value of x=a in the given function. \nonumber \], Since \(\displaystyle \lim_{xa}f(x)=,\) we know that \(\displaystyle \lim_{xa}\ln(f(x))=\). Limits of this form are classified as indeterminate forms of type \(/\). Rule 65. Injunctions and Restraining Orders | Federal Rules of which is another \(\frac00\) form. Rules Governing the Food Protection and Sanitation of Food \nonumber \], \(\dfrac{d}{dx}\big(\tan x\big)=\sec^2x\). }\), Evaluate \(\lim\limits_{x\rightarrow 0}\dfrac{\sin(x^3+3x^2)}{\sin^2x}\text{. \nonumber \]. Usually, it is best to find a common factor or find a common denominator to convert it into a form where }\), Evaluate \(\lim\limits_{x\rightarrow 0} \dfrac{xe^x}{\tan (3x)}\text{. Let, \[\ln y=\ln(f(x)^{g(x)})=g(x)\ln(f(x)). The first term converges to 0 (by the squeeze theorem), but the second term \(\cos(1/x)\) just oscillates wildly between \(\pm 1\text{. }\) We can use a little algebra to manipulate this into either a \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) form: \begin{align*} \lim_{x \to a} \frac{f(x)}{1/g(x)} && \lim_{x \to a} \frac{g(x)}{1/f(x)} \end{align*}, \begin{gather*} \lim_{x\to 0^+} x \cdot \log x \end{gather*}, Here the function \(f(x)=x\) goes to zero, while \(g(x)=\log x\) goes to \(-\infty\text{. In this rule we find the derivative of the numerator and denominator separately and then substitute the values to find the value for the limit.In case we still get an The expression \(0\) is considered indeterminate because we cannot determine without further analysis the exact behavior of the product \(f(x)g(x)\) as \(x\). An indeterminate form can take any value ( or no value ). For example, let \(n\) be a positive integer and consider. In this case, applying LHpitals rule, we would obtain, \[\lim_{x0^+}\sin x\ln x=\lim_{x0^+}\dfrac{\sin x}{1/\ln x}=\lim_{x0^+}\dfrac{\cos x}{1/(x(\ln x)^2)}=\lim_{x0^+}(x(\ln x)^2\cos x).\nonumber \], Unfortunately, we not only have another expression involving the indeterminate form \(0,\) but the new limit is even more complicated to evaluate than the one with which we started. \begin{gather*} \lim_{x\to 0} \frac{\sin x}{x} \end{gather*}, \begin{align*} \lim_{x\to 0} \sin x &= 0\\ \lim_{x\to 0} x &= 0 \end{align*}, \begin{align*} f(x)&= \sin x & f'(x) & =\cos x & \text{and} && f'(0)=1\\ g(x) &= x & g'(x) & = 1 & \text{and} && g'(0)=1 \end{align*}, \begin{align*} \lim_{x\to 0} \frac{\sin x}{x} &= \frac{f'(0)}{g'(0)} = \frac{1}{1} = 1. If we try to do so, we get, At which point we would conclude erroneously that, \[\lim_{x1}\dfrac{x^2+5}{3x+4}=\lim_{x1}\dfrac{2x}{3}=\dfrac{2}{3}. \nonumber \], \[\lim_{x}(f(x)g(x))=\lim_{x}(3x^23x^25)=5. \nonumber \], We now evaluate \(\displaystyle\lim_{x0^+}\sin x\ln x.\) Since \(\displaystyle\lim_{x0^+}\sin x=0\) and \(\displaystyle\lim_{x0^+}\ln x=\), we have the indeterminate form \(0\). 0 following examples: x 2 4 (x + 2)(x 2) Example 1: lim = lim . Applying l'Hpital's rule again gives: \begin{align*} \lim_{x\rightarrow 0}\underbrace{ \frac{x\cos x -\sin x}{2x^2\sin x} }_{{\mathrm{num}\rightarrow 0} \\{\mathrm{den}\rightarrow 0}} &=\lim_{x\rightarrow 0} \frac{\cos x -x\sin x-\cos x}{4x\sin x+2x^2\cos x}\\ &=-\lim_{x\rightarrow 0} \frac{x\sin x}{4x\sin x+2x^2\cos x} =-\lim_{x\rightarrow 0} \frac{\sin x}{4\sin x+2x\cos x} \end{align*}. }\), Evaluate \(\lim\limits_{x \to 0}\sqrt[x^2]{\cos x}\text{. This result also holds if we are considering one-sided limits, or if \(a=\) or \(a=.\). \nonumber \], Now as \(x0^+, \csc^2x\). }\), Evaluate \(\displaystyle\lim_{x \to \infty}\sqrt{2x^2+1}-\sqrt{x^2+x}\text{. Calculus I - L'Hospital's Rule and Indeterminate Forms Compare the growth rates of \(x^{100}\) and \(2^x\). Identify indeterminate forms produced by quotients, products, subtractions, and powers, and apply LHpitals rule in each case.
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