indeterminate forms and l'hospital's rule

a. Example 3.7.4 Compute \(\displaystyle \lim_{x\to 0} \frac{\sin(x)}{\sin(2x)}\). Evaluate \(\lim\limits_{x\rightarrow 0}\dfrac{e^{k\sin(x^2)}-(1+2x^2)}{x^4}\text{,}\) where \(k\) is a constant. Customer Service Center: 1-800-662-7030 Visit RelayNC for information }\) Then the GMVT (Theorem 3.4.38) tells us that for \(x\in (a,b]\), \begin{align*} \frac{f(x)}{g(x)} = \frac{f(x)-f(a)}{g(x)-g(a)} &= \frac{f'(c)}{g'(c)} \end{align*}, where \(c \in (a,x)\text{. \nonumber \]. 0Then f(x) = lim \begin{gather*} \lim_{x\to a^+} \frac{f(x)}{g(x)} \end{gather*}, \begin{gather*} \lim_{x\to a^+} f(x) = \lim_{x\to a^+} g(x) = 0 \end{gather*}, For simplicity, we also assume that \(f(a)=g(a)=0\text{. }\) If we rewrite this as the fraction, \begin{align*} x \cdot \log x &= \frac{\log x}{1/x} \end{align*}. Rule 65. For example, let \(n\) be a positive integer and consider. f(x) g(x) = 0 0. or. For example, suppose the exponent n in the function \(f(x)=3x^n\) is \(n=3\), then, \[\lim_{x}(f(x)g(x))=\lim_{x}(3x^33x^25)=. \nonumber \], We need to evaluate \(\displaystyle \lim_{x}\dfrac{\ln x}{x}\). Applying l'Hpital's rule gives, \begin{gather*} \lim_{x\rightarrow 0}\underbrace{\frac{a\log(1+x)}{x}}_{ {\mathrm{num}\rightarrow 0}\\{\mathrm{den}\rightarrow 0}} =\lim_{x\rightarrow 0}\underbrace{\frac{\frac{a}{1+x}}{1}}_{{\mathrm{num}\rightarrow a} \\{\mathrm{den}\rightarrow 1}} =a \end{gather*}, Since \((1 + x)^{a/x} = \exp\left[\log\Big((1 + x)^{a/x}\Big)\right]\) and the exponential function is continuous, our original limit is \(e^a\text{.}\). LHospitals Rule works great on the two indeterminate forms 0/0 and \({{ \pm \,\infty }}/{{ \pm \,\infty }}\;\). Note that around that time l'Hpital's name was commonly spelled l'Hospital, but the spelling of silent s in French was changed subsequently; many texts spell his name l'Hospital. Since the least common denominator is \(x^2\tan x,\) we have. WebDenition 1.2. \end{align*}, If we attempt to apply l'Hptial's rule then we have \(g'(x)=1\) and, \begin{align*} f'(x)&=2x\sin\frac{1}{x} -\cos\frac{1}{x}\\ \end{align*}, However, this limit does not exist. If you find yourself in Paris, you can hunt along Boulevard de l'Hpital for older street signs carved into the sides of buildings which spell it l'Hospital though arguably there are better things to do there. This page titled 3.7: L'Hpital's Rule and Indeterminate Forms is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. Since, \[\lim_{x}\frac{2x}{e^x}=\lim_{x}\frac{2}{e^x}=0, \nonumber \], \[\lim_{x}\dfrac{x^2}{e^x}=0. WebIndeterminate Forms and LHospitals Rule Review: Limits Involving Quotients As we have seen one of the recurring problems in this course is nding the limit of the quotient oftwo functions. \nonumber \], Now \(\displaystyle\lim_{x0^+}\sin^2x=0\) and \(\displaystyle\lim_{x0^+}-x=0\), so we apply LHpitals rule again. Next we realize why these are indeterminate forms and then understand how to use LHpitals rule in these cases. which is yet another \(\frac00\) form. Again, note that we are not actually dividing \(\) by \(\). \nonumber \], Evaluate \[\lim_{x}x^{1/x}. &=\lim_{xa}\dfrac{\dfrac{f(x)f(a)}{xa}}{\dfrac{g(x)g(a)}{xa}} & & \text{Multiply numerator and denominator by}\, \frac{1}{x-a} \\[4pt] WebIndeterminate Form If the continuous 0/0 functions are both zero atx= g(x) a, f(x)and then cannot develop such lim lim x be found by general m its. WebWhether you are in need of regular primary care and checkups or surgical services and recovery, our physicians and medical professionals have the technology, knowledge and Note however that it is in the following indeterminate Because the limits of the numerator and denominator are not both zero and are not both infinite, we cannot apply LHpitals rule. }\) All we can conclude from this is. \nonumber \], \[\lim_{x0^+}\left(\dfrac{1}{x^2}\dfrac{1}{\tan x}\right)=. If we write, we see that \(\ln x\) as \(x0^+\) and \(\dfrac{1}{x}\) as \(x0^+\). \begin{gather*} \lim_{x\rightarrow +\infty}x^{\frac{1}{x}} \end{gather*}, the base, \(x\text{,}\) goes to infinity and the exponent, \(\frac{1}{x}\text{,}\) goes to zero. The following video shows how to manipulate indeterminate differences to make them WebGet an indeterminate of the form (this is not necessarily zero!). 0 0 Ex. Depending on whether \(f(x)\) grows faster, \(g(x)\) grows faster, or they grow at the same rate, as we see next, anything can happen in this limit. To apply the rule we must first check the limits of the derivatives. We can also use LHpitals rule to evaluate limits of quotients \(\dfrac{f(x)}{g(x)}\) in which \(f(x)\) and \(g(x)\). Legal. In fact, we already found, in Example 3.7.15, that, \begin{gather*} \lim_{x\rightarrow 0+} x\log x=0 \end{gather*}, Since the exponential is a continuous function, \begin{gather*} \lim_{x\rightarrow 0+}x^x =\lim_{x\rightarrow 0+}\exp\big(x\log x\big) =\exp\Big(\lim_{x\rightarrow 0+}x\log x\Big) =e^0 =1 \end{gather*}. }\) To see this note that since derivative \(f'(a)\) exists, we know that the limit. &=\frac{\displaystyle \lim_{xa}f(x)}{\displaystyle \lim_{xa}g(x)} & & \text{By the continuity of}\, f \,\text{and}\, g \\[4pt] \begin{align*} \lim\limits_{x\rightarrow a}f(x)&=0 & \text{but}&& \lim\limits_{x\rightarrow a}g(x)&\ne 0 \end{align*}, \begin{align*} \lim\limits_{x\rightarrow a}\frac{f(x)}{g(x)} && \text{need not be the same as} && \frac{f'(a)}{g'(a)} \text{ or } \lim\limits_{x\rightarrow a}\frac{f'(x)}{g'(x)}. If, \begin{align*} \lim_{x\to a} f(x) &= 0 & \text{and } && \lim_{x\to a} g(x) &= 0 \end{align*}, \begin{gather*} \lim_{x\to a} \dfrac{f(x)}{g(x)} \end{gather*}. Earlier in the chapter we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of x in the denominator. WebL'H^opital's Rule and Indeterminate Forms Created by Tynan Lazarus October 29, 2018 Now that we have the power of the derivative, we can use it as a way to compute limits A researcher writes that the algorithm will terminate in roughly at most \(A(n)=5n^4\) steps. 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*}. Describe the relative growth rates of functions. Remark: this is a very common kind of approximation. Section 4.10 : L'Hospital's Rule and Indeterminate Forms. Let, \[\ln y=\ln(f(x)^{g(x)})=g(x)\ln(f(x)). 0 following examples: x 2 4 (x + 2)(x 2) Example 1: lim = lim . http://www.ncjj.org/About/History.aspx \begin{gather*} \lim_{x\rightarrow 0}(1+x)^{\frac{a}{x}} \qquad\text{with the constant } a\ne 0 \end{gather*}. The court may These expressions are not real numbers. 1.lim x!2+ [ln(3x2 12) ln(x4 16)] 2.lim x!0+ 2x+ 1 sinx 1 x 3 While we can't divide by zero, we can still meaningfully ask what happens to a ratio f ( x) g ( x) when both f ( x) }\), Evaluate \(\lim\limits_{x\rightarrow\infty}x^2e^{-x}\text{. Then, if necessary, we apply LHospitals Rule. / . x . WeakLHospitals Strong ApplyingLHospitals Other Indeterminate Forms In the last example, Example 3.7.17, we converted an \(\infty-\infty\) indeterminate form into a \(\frac{0}{0}\) indeterminate form by exploiting the fact that the two terms, \(\sec x\) and \(\tan x\text{,}\) in the \(\infty-\infty\) indeterminate form shared a common denominator, namely \(\cos x\text{. We have just shown that the logarithm of our original limit is \(-\frac{1}{6}\text{. \(f(x)=\dfrac{1}{(x^n+1)}\) and \(g(x)=3x^2\). This can be further transformed into a \(\frac00\) or \(\frac\infty\infty\) form: \begin{align*} \log\left( \lim_{x\to a} f(x)^{g(x)} \right) &=\lim_{x\to a} \log\left( f(x) \right) \cdot g(x)\\ &= \lim_{x\to a} \frac{\log\left( f(x) \right)}{1/g(x)}. But if we consider the logarithm then we have, \begin{gather*} \log x^x = x\log x \end{gather*}, which is a \(0\cdot\infty\) indeterminate form, which we already know how to treat. Lhopitals rule states that if f(x) & g(x) are differentiable functions and d/dx [g(x)] 0 on an open interval. WebL'Hospital's Rule and Indeterminate Forms What does 0 0 mean? }\) In the real world that will, of course, almost never happen. Like the \(1^\infty\) form, this can be treated by considering its logarithm. \nonumber \], \[\lim_{xa}[\ln(y)]=\lim_{xa}[g(x)\ln(f(x))]. In this example, we shall apply L'Hpital's rule twice before getting the answer. \nonumber \], \[\lim_{x0^+}\dfrac{\ln x}{\cot x}=0. WebThe indeterminate forms are expressions of the form of 0/0, 0 0, 0 x (), - , 1 , 0, and / after computing limits. \nonumber \], \[\lim_{x}(f(x)g(x))=\lim_{x}(3x3x^25)=. Let \(n\) be a positive integer and let \(f(x)=3x^n\) and \(g(x)=3x^2+5\). Example 3.7.3 Find \(\displaystyle \lim_{x\to 0} \frac{\sin x}{x}\). }\), Evaluate \(\lim\limits_{x\rightarrow 0} \dfrac{xe^x}{\tan (3x)}\text{. We convert it into a \(\frac{0}{0}\) indeterminate form simply by putting the two fractions, \(\frac{1}{x}\) and \(\frac{1}{\log(1+x)}\) over a common denominator. The requirement that \(G\ne 0\) is critical we explored this in Example 1.4.7. We call this one of the indeterminate forms, of type \(\dfrac{0}{0}\). But if we take logarithms, \begin{gather*} \log x^{\frac{1}{x}} =\frac{\log x}{x} \end{gather*}. This result also holds if we are considering one-sided limits, or if \(a=\) or \(a=.\). &=\lim_{xa}\dfrac{f(x)}{g(x)}. 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! Evaluate \(\displaystyle\lim_{x0^+}\dfrac{\cos x}{x}\) by other means. \nonumber \], \(\dfrac{d}{dx}\big(\ln x\big)=\dfrac{1}{x}\). Please reread that example. \nonumber \], Since \(\displaystyle \lim_{x}2x=\) and \(\displaystyle \lim_{x}e^x=\), we can apply LHpitals rule again. WebIndeterminate Forms and LHospitals Rule Learning objectives LHospitals Rule is a powerful technique for computing limits of large classes of functions. \nonumber \]. L Hospitals Rule Suppose that f and g are differentiable functions and g^\prime { (x)} \neq 0 on an open interval I that contains a (except possibly at a ). 2 = lim (x + 2) = 2 + 2 = 4 . Here is another, much more complicated, example, where it doesn't happen. We have previously studied limits with the indeterminate form as shown in the . It is important to remember, however, that to apply LHpitals rule to a quotient \(\dfrac{f(x)}{g(x)}\), it is essential that the limit of \(\dfrac{f(x)}{g(x)}\) be of the form \(\dfrac{0}{0}\) or \(/\). For example, \(f(x)=x^2\) and \(g(x)=x^3\) both approach infinity as \(x\). }\) 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*}. }\) 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. Therefore, the first term in the denominator is approaching zero and the second term is getting really large. }\), By comparison, if we multiply by the conjugate we have, \begin{align*} \sqrt{x^2\!+\!4x}-\sqrt{x^2\!-\!3x} &= \left( \sqrt{x^2\!+\!4x}-\sqrt{x^2\!-\!3x}\right) \cdot \frac{\sqrt{x^2\!+\!4x}+\sqrt{x^2\!-\!3x}}{\sqrt{x^2\!+\!4x}+\sqrt{x^2\!-\!3x}}\\ &= \frac{ x^2+4x - (x^2-3x)}{\sqrt{x^2+4x}+\sqrt{x^2-3x}}\\ &= \frac{ 7x}{\sqrt{x^2+4x}+\sqrt{x^2-3x}}\\ &= \frac{7}{\sqrt{1+4/x}+\sqrt{1-3/x}} \qquad \text{assuming } x \gt 0 \end{align*}. Therefore, \[\begin{align*} \lim_{xa}\dfrac{f(x)}{g(x)} &=\lim_{xa}\dfrac{f(x)f(a)}{g(x)g(a)} & & \text{Since} \, f(a)=0=g(a) \\[4pt] Although the values of both functions become arbitrarily large as the values of \(x\) become sufficiently large, sometimes one function is growing more quickly than the other. We obtain, \[\lim_{x}\dfrac{e^{1/x}1}{\dfrac{1}{x}}=\lim_{x}\dfrac{e^{1/x}(\tfrac{1}{x^2})}{\left(\frac{1}{x^2}\right)}=\lim_{x}e^{1/x}=e^0=1. To deal with this type of limit, we use algebra to introduce a quotient. There are quite a number of mathematical tools for evaluating such indeterminate forms Taylor series for example. This was quite a complicated example. \nonumber \], \[\lim_{x}(f(x)g(x))=\lim_{x}(3x^23x^25)=5. We know, from Theorem 1.4.3 on the arithmetic of limits, that if, \begin{align*} \lim_{x\rightarrow a}f(x) &= F & \lim_{x\rightarrow a}g(x) &= G\\ \end{align*}, \begin{align*} \lim_{x\rightarrow a}\frac{f(x)}{g(x)} &= \frac{F}{G} \end{align*}. it is not difficult to show that \(e^x\) grows more rapidly than \(x^p\) for any \(p>0\). Explain why we cannot apply LHpitals rule to evaluate \(\displaystyle\lim_{x0^+}\dfrac{\cos x}{x}\). WebFederal Rules of Civil Procedure. Injunctions and Restraining Orders. Recall: (QuotientLaw) f(x) limlimf(x) = x!ag(x)x!a limg(x) x!a if limg(x)6= 0 x!a limg(x)6= 0 x!a limg(x) = 0 x!a Indeterminate Quotients

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