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Indeterminate form
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<h2 id="some-examples-and-non-examples">Some examples and non-examples</h2> <h3>Indeterminate form 0/0</h3> <p class="note">"0/0" redirects here. For the symbol, see <a href="/facts/Percent_sign/pLdHNKar">Percent sign</a>. For 0 divided by 0, see <a href="/facts/Division_by_zero/wXK7t2Ge">Division by zero</a>.</p> <p>The indeterminate form 0 / 0 {\displaystyle 0/0} is particularly common in <a href="/facts/Calculus/MEmUTYoz">calculus</a>, because it often arises in the evaluation of <a href="/facts/Derivative/7Ito70Dh">derivatives</a> using their definition in terms of limit. </p><p>As mentioned above, </p> lim x → 0 x x = 1 , {\displaystyle \lim _{x\to 0}{\frac {x}{x}}=1,\qquad } (see fig. 1) <p>while </p> lim x → 0 x 2 x = 0 , {\displaystyle \lim _{x\to 0}{\frac {x^{2}}{x}}=0,\qquad } (see fig. 2) <p>This is enough to show that 0 / 0 {\displaystyle 0/0} is an indeterminate form. Other examples with this indeterminate form include </p> lim x → 0 sin ( x ) x = 1 , {\displaystyle \lim _{x\to 0}{\frac {\sin(x)}{x}}=1,\qquad } (see fig. 3) <p>and </p> lim x → 49 x − 49 x − 7 = 14 , {\displaystyle \lim _{x\to 49}{\frac {x-49}{{\sqrt {x}}\,-7}}=14,\qquad } (see fig. 4) <p>Direct substitution of the number that <i> x {\displaystyle x} </i> approaches into any of these expressions shows that these are examples correspond to the indeterminate form 0 / 0 {\displaystyle 0/0} , but these limits can assume many different values. Any desired value a {\displaystyle a} can be obtained for this indeterminate form as follows: </p> lim x → 0 a x x = a . {\displaystyle \lim _{x\to 0}{\frac {ax}{x}}=a.\qquad } (see fig. 5) <p>The value ∞ {\displaystyle \infty } can also be obtained (in the sense of divergence to infinity): </p> lim x → 0 x x 3 = ∞ . {\displaystyle \lim _{x\to 0}{\frac {x}{x^{3}}}=\infty .\qquad } (see fig. 6) <h3>Indeterminate form 00</h3> <p class="note">Main article: <a href="/facts/Zero_to_the_power_of_zero/pxbWkuFg">Zero to the power of zero</a></p> <p>The following limits illustrate that the expression 0 0 {\displaystyle 0^{0}} is an indeterminate form: lim x → 0 + x 0 = 1 , lim x → 0 + 0 x = 0. {\displaystyle {\begin{aligned}\lim _{x\to 0^{+}}x^{0}&=1,\\\lim _{x\to 0^{+}}0^{x}&=0.\end{aligned}}} </p><p>Thus, in general, knowing that lim x → c f ( x ) = 0 {\displaystyle \textstyle \lim _{x\to c}f(x)\;=\;0} and lim x → c g ( x ) = 0 {\displaystyle \textstyle \lim _{x\to c}g(x)\;=\;0} is not sufficient to evaluate the limit lim x → c f ( x ) g ( x ) . {\displaystyle \lim _{x\to c}f(x)^{g(x)}.} </p><p>If the functions f {\displaystyle f} and g {\displaystyle g} are <a href="/facts/Analytic_function/JhL3z8FP">analytic</a> at c {\displaystyle c} , and f {\displaystyle f} is positive for x {\displaystyle x} sufficiently close (but not equal) to c {\displaystyle c} , then the limit of f ( x ) g ( x ) {\displaystyle f(x)^{g(x)}} will be 1 {\displaystyle 1} .<a class="footnote-ref" id="fnref:3" href="#fn:3"><sup>3</sup></a> Otherwise, use the transformation in the table below to evaluate the limit. </p> <h3>Expressions that are not indeterminate forms</h3> <p>The expression 1 / 0 {\displaystyle 1/0} is not commonly regarded as an indeterminate form, because if the limit of f / g {\displaystyle f/g} exists then there is no ambiguity as to its value, as it always diverges. Specifically, if f {\displaystyle f} approaches 1 {\displaystyle 1} and g {\displaystyle g} approaches 0 , {\displaystyle 0,} then f {\displaystyle f} and g {\displaystyle g} may be chosen so that: </p> <ol><li> f / g {\displaystyle f/g} approaches + ∞ {\displaystyle +\infty } </li> <li> f / g {\displaystyle f/g} approaches − ∞ {\displaystyle -\infty } </li> <li>The limit fails to exist.</li></ol> <p>In each case the absolute value | f / g | {\displaystyle |f/g|} approaches + ∞ {\displaystyle +\infty } , and so the quotient f / g {\displaystyle f/g} must diverge, in the sense of the <a href="/facts/Extended_real_number/65Gs7QS0">extended real numbers</a> (in the framework of the <a href="/facts/Projectively_extended_real_line/9epfpLw2">projectively extended real line</a>, the limit is the <a href="/facts/Point_at_infinity/GAuS0wHQ">unsigned infinity</a> ∞ {\displaystyle \infty } in all three cases<a class="footnote-ref" id="fnref:4" href="#fn:4"><sup>4</sup></a>). Similarly, any expression of the form a / 0 {\displaystyle a/0} with a ≠ 0 {\displaystyle a\neq 0} (including a = + ∞ {\displaystyle a=+\infty } and a = − ∞ {\displaystyle a=-\infty } ) is not an indeterminate form, since a quotient giving rise to such an expression will always diverge. </p><p>The expression 0 ∞ {\displaystyle 0^{\infty }} is not an indeterminate form. The expression 0 + ∞ {\displaystyle 0^{+\infty }} obtained from considering lim x → c f ( x ) g ( x ) {\displaystyle \lim _{x\to c}f(x)^{g(x)}} gives the limit 0 , {\displaystyle 0,} provided that f ( x ) {\displaystyle f(x)} remains nonnegative as x {\displaystyle x} approaches c {\displaystyle c} . The expression 0 − ∞ {\displaystyle 0^{-\infty }} is similarly equivalent to 1 / 0 {\displaystyle 1/0} ; if f ( x ) > 0 {\displaystyle f(x)>0} as x {\displaystyle x} approaches c {\displaystyle c} , the limit comes out as + ∞ {\displaystyle +\infty } . </p><p>To see why, let L = lim x → c f ( x ) g ( x ) , {\displaystyle L=\lim _{x\to c}f(x)^{g(x)},} where lim x → c f ( x ) = 0 , {\displaystyle \lim _{x\to c}{f(x)}=0,} and lim x → c g ( x ) = ∞ . {\displaystyle \lim _{x\to c}{g(x)}=\infty .} By taking the natural logarithm of both sides and using lim x → c ln f ( x ) = − ∞ , {\displaystyle \lim _{x\to c}\ln {f(x)}=-\infty ,} we get that ln L = lim x → c ( g ( x ) × ln f ( x ) ) = ∞ × − ∞ = − ∞ , {\displaystyle \ln L=\lim _{x\to c}({g(x)}\times \ln {f(x)})=\infty \times {-\infty }=-\infty ,} which means that L = e − ∞ = 0. {\displaystyle L={e}^{-\infty }=0.} </p> <h2 id="evaluating-indeterminate-forms">Evaluating indeterminate forms</h2> <p>The adjective <i>indeterminate</i> does <i>not</i> imply that the limit does not exist, as many of the examples above show. In many cases, algebraic elimination, <a href="/facts/L%2527H%25C3%25B4pital%2527s_rule/pCXK75Io">L'Hôpital's rule</a>, or other methods can be used to manipulate the expression so that the limit can be evaluated. </p> <h3>Equivalent infinitesimal</h3> <p>When two variables α {\displaystyle \alpha } and β {\displaystyle \beta } converge to zero at the same limit point and lim β α = 1 {\displaystyle \textstyle \lim {\frac {\beta }{\alpha }}=1} , they are called <i>equivalent infinitesimal</i> (equiv. α ∼ β {\displaystyle \alpha \sim \beta } ). </p><p>Moreover, if variables α ′ {\displaystyle \alpha '} and β ′ {\displaystyle \beta '} are such that α ∼ α ′ {\displaystyle \alpha \sim \alpha '} and β ∼ β ′ {\displaystyle \beta \sim \beta '} , then: </p> lim β α = lim β ′ α ′ {\displaystyle \lim {\frac {\beta }{\alpha }}=\lim {\frac {\beta '}{\alpha '}}} <p>Here is a brief proof: </p><p>Suppose there are two equivalent infinitesimals α ∼ α ′ {\displaystyle \alpha \sim \alpha '} and β ∼ β ′ {\displaystyle \beta \sim \beta '} . </p><p> lim β α = lim β β ′ α ′ β ′ α ′ α = lim β β ′ lim α ′ α lim β ′ α ′ = lim β ′ α ′ {\displaystyle \lim {\frac {\beta }{\alpha }}=\lim {\frac {\beta \beta '\alpha '}{\beta '\alpha '\alpha }}=\lim {\frac {\beta }{\beta '}}\lim {\frac {\alpha '}{\alpha }}\lim {\frac {\beta '}{\alpha '}}=\lim {\frac {\beta '}{\alpha '}}} </p><p>For the evaluation of the indeterminate form 0 / 0 {\displaystyle 0/0} , one can make use of the following facts about equivalent <a href="/facts/Infinitesimal/JKcTMIgD">infinitesimals</a> (e.g., x ∼ sin x {\displaystyle x\sim \sin x} if <i>x</i> becomes closer to zero):<a class="footnote-ref" id="fnref:5" href="#fn:5"><sup>5</sup></a> </p> x ∼ sin x , {\displaystyle x\sim \sin x,} x ∼ arcsin x , {\displaystyle x\sim \arcsin x,} x ∼ sinh x , {\displaystyle x\sim \sinh x,} x ∼ tan x , {\displaystyle x\sim \tan x,} x ∼ arctan x , {\displaystyle x\sim \arctan x,} x ∼ ln ( 1 + x ) , {\displaystyle x\sim \ln(1+x),} 1 − cos x ∼ x 2 2 , {\displaystyle 1-\cos x\sim {\frac {x^{2}}{2}},} cosh x − 1 ∼ x 2 2 , {\displaystyle \cosh x-1\sim {\frac {x^{2}}{2}},} a x − 1 ∼ x ln a , {\displaystyle a^{x}-1\sim x\ln a,} e x − 1 ∼ x , {\displaystyle e^{x}-1\sim x,} ( 1 + x ) a − 1 ∼ a x . {\displaystyle (1+x)^{a}-1\sim ax.} <p>For example: </p><p> lim x → 0 1 x 3 [ ( 2 + cos x 3 ) x − 1 ] = lim x → 0 e x ln 2 + cos x 3 − 1 x 3 = lim x → 0 1 x 2 ln 2 + cos x 3 = lim x → 0 1 x 2 ln ( cos x − 1 3 + 1 ) = lim x → 0 cos x − 1 3 x 2 = lim x → 0 − x 2 6 x 2 = − 1 6 {\displaystyle {\begin{aligned}\lim _{x\to 0}{\frac {1}{x^{3}}}\left[\left({\frac {2+\cos x}{3}}\right)^{x}-1\right]&=\lim _{x\to 0}{\frac {e^{x\ln {\frac {2+\cos x}{3}}}-1}{x^{3}}}\\&=\lim _{x\to 0}{\frac {1}{x^{2}}}\ln {\frac {2+\cos x}{3}}\\&=\lim _{x\to 0}{\frac {1}{x^{2}}}\ln \left({\frac {\cos x-1}{3}}+1\right)\\&=\lim _{x\to 0}{\frac {\cos x-1}{3x^{2}}}\\&=\lim _{x\to 0}-{\frac {x^{2}}{6x^{2}}}\\&=-{\frac {1}{6}}\end{aligned}}} </p><p>In the 2nd equality, e y − 1 ∼ y {\displaystyle e^{y}-1\sim y} where y = x ln 2 + cos x 3 {\displaystyle y=x\ln {2+\cos x \over 3}} as <i>y</i> become closer to 0 is used, and y ∼ ln ( 1 + y ) {\displaystyle y\sim \ln {(1+y)}} where y = cos x − 1 3 {\displaystyle y={{\cos x-1} \over 3}} is used in the 4th equality, and 1 − cos x ∼ x 2 2 {\displaystyle 1-\cos x\sim {x^{2} \over 2}} is used in the 5th equality. </p> <h3>L'Hôpital's rule</h3> <p class="note">Main article: <a href="/facts/L%2527H%25C3%25B4pital%2527s_rule/pCXK75Io">L'Hôpital's rule</a></p> <p>L'Hôpital's rule is a general method for evaluating the indeterminate forms 0 / 0 {\displaystyle 0/0} and ∞ / ∞ {\displaystyle \infty /\infty } . This rule states that (under appropriate conditions) </p> lim x → c f ( x ) g ( x ) = lim x → c f ′ ( x ) g ′ ( x ) , {\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {f'(x)}{g'(x)}},} <p>where f ′ {\displaystyle f'} and g ′ {\displaystyle g'} are the <a href="/facts/Derivative_(calculus)/7Ito70Dh">derivatives</a> of f {\displaystyle f} and g {\displaystyle g} . (Note that this rule does <i>not</i> apply to expressions ∞ / 0 {\displaystyle \infty /0} , 1 / 0 {\displaystyle 1/0} , and so on, as these expressions are not indeterminate forms.) These derivatives will allow one to perform algebraic simplification and eventually evaluate the limit. </p><p>L'Hôpital's rule can also be applied to other indeterminate forms, using first an appropriate algebraic transformation. For example, to evaluate the form 00: </p> ln lim x → c f ( x ) g ( x ) = lim x → c ln f ( x ) 1 / g ( x ) . {\displaystyle \ln \lim _{x\to c}f(x)^{g(x)}=\lim _{x\to c}{\frac {\ln f(x)}{1/g(x)}}.} <p>The right-hand side is of the form ∞ / ∞ {\displaystyle \infty /\infty } , so L'Hôpital's rule applies to it. Note that this equation is valid (as long as the right-hand side is defined) because the <a href="/facts/Natural_logarithm/PnreKEzI">natural logarithm</a> (ln) is a <a href="/facts/Continuous_function/sKbl02pB">continuous function</a>; it is irrelevant how well-behaved f {\displaystyle f} and g {\displaystyle g} may (or may not) be as long as f {\displaystyle f} is asymptotically positive. (the domain of logarithms is the set of all positive real numbers.) </p><p>Although L'Hôpital's rule applies to both 0 / 0 {\displaystyle 0/0} and ∞ / ∞ {\displaystyle \infty /\infty } , one of these forms may be more useful than the other in a particular case (because of the possibility of algebraic simplification afterwards). One can change between these forms by transforming f / g {\displaystyle f/g} to ( 1 / g ) / ( 1 / f ) {\displaystyle (1/g)/(1/f)} . </p> <h2 id="list-of-indeterminate-forms">List of indeterminate forms</h2> <p>The following table lists the most common indeterminate forms and the transformations for applying l'Hôpital's rule. </p> <table><tbody><tr><th>Indeterminate form</th><th>Conditions</th><th>Transformation to 0 / 0 {\displaystyle 0/0} </th><th>Transformation to ∞ / ∞ {\displaystyle \infty /\infty } </th></tr><tr><td> 0 {\displaystyle 0} / 0 {\displaystyle 0} </td><td> lim x → c f ( x ) = 0 , lim x → c g ( x ) = 0 {\displaystyle \lim _{x\to c}f(x)=0,\ \lim _{x\to c}g(x)=0\!} </td><td>—</td><td> lim x → c f ( x ) g ( x ) = lim x → c 1 / g ( x ) 1 / f ( x ) {\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {1/g(x)}{1/f(x)}}\!} </td></tr><tr><td> ∞ {\displaystyle \infty } / ∞ {\displaystyle \infty } </td><td> lim x → c f ( x ) = ∞ , lim x → c g ( x ) = ∞ {\displaystyle \lim _{x\to c}f(x)=\infty ,\ \lim _{x\to c}g(x)=\infty \!} </td><td> lim x → c f ( x ) g ( x ) = lim x → c 1 / g ( x ) 1 / f ( x ) {\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {1/g(x)}{1/f(x)}}\!} </td><td>—</td></tr><tr><td> 0 ⋅ ∞ {\displaystyle 0\cdot \infty } </td><td> lim x → c f ( x ) = 0 , lim x → c g ( x ) = ∞ {\displaystyle \lim _{x\to c}f(x)=0,\ \lim _{x\to c}g(x)=\infty \!} </td><td> lim x → c f ( x ) g ( x ) = lim x → c f ( x ) 1 / g ( x ) {\displaystyle \lim _{x\to c}f(x)g(x)=\lim _{x\to c}{\frac {f(x)}{1/g(x)}}\!} </td><td> lim x → c f ( x ) g ( x ) = lim x → c g ( x ) 1 / f ( x ) {\displaystyle \lim _{x\to c}f(x)g(x)=\lim _{x\to c}{\frac {g(x)}{1/f(x)}}\!} </td></tr><tr><td> ∞ − ∞ {\displaystyle \infty -\infty } </td><td> lim x → c f ( x ) = ∞ , lim x → c g ( x ) = ∞ {\displaystyle \lim _{x\to c}f(x)=\infty ,\ \lim _{x\to c}g(x)=\infty \!} </td><td> lim x → c ( f ( x ) − g ( x ) ) = lim x → c 1 / g ( x ) − 1 / f ( x ) 1 / ( f ( x ) g ( x ) ) {\displaystyle \lim _{x\to c}(f(x)-g(x))=\lim _{x\to c}{\frac {1/g(x)-1/f(x)}{1/(f(x)g(x))}}\!} </td><td> lim x → c ( f ( x ) − g ( x ) ) = ln lim x → c e f ( x ) e g ( x ) {\displaystyle \lim _{x\to c}(f(x)-g(x))=\ln \lim _{x\to c}{\frac {e^{f(x)}}{e^{g(x)}}}\!} </td></tr><tr><td> 0 0 {\displaystyle 0^{0}} </td><td> lim x → c f ( x ) = 0 + , lim x → c g ( x ) = 0 {\displaystyle \lim _{x\to c}f(x)=0^{+},\lim _{x\to c}g(x)=0\!} </td><td> lim x → c f ( x ) g ( x ) = exp lim x → c g ( x ) 1 / ln f ( x ) {\displaystyle \lim _{x\to c}f(x)^{g(x)}=\exp \lim _{x\to c}{\frac {g(x)}{1/\ln f(x)}}\!} </td><td> lim x → c f ( x ) g ( x ) = exp lim x → c ln f ( x ) 1 / g ( x ) {\displaystyle \lim _{x\to c}f(x)^{g(x)}=\exp \lim _{x\to c}{\frac {\ln f(x)}{1/g(x)}}\!} </td></tr><tr><td> 1 ∞ {\displaystyle 1^{\infty }} </td><td> lim x → c f ( x ) = 1 , lim x → c g ( x ) = ∞ {\displaystyle \lim _{x\to c}f(x)=1,\ \lim _{x\to c}g(x)=\infty \!} </td><td> lim x → c f ( x ) g ( x ) = exp lim x → c ln f ( x ) 1 / g ( x ) {\displaystyle \lim _{x\to c}f(x)^{g(x)}=\exp \lim _{x\to c}{\frac {\ln f(x)}{1/g(x)}}\!} </td><td> lim x → c f ( x ) g ( x ) = exp lim x → c g ( x ) 1 / ln f ( x ) {\displaystyle \lim _{x\to c}f(x)^{g(x)}=\exp \lim _{x\to c}{\frac {g(x)}{1/\ln f(x)}}\!} </td></tr><tr><td> ∞ 0 {\displaystyle \infty ^{0}} </td><td> lim x → c f ( x ) = ∞ , lim x → c g ( x ) = 0 {\displaystyle \lim _{x\to c}f(x)=\infty ,\ \lim _{x\to c}g(x)=0\!} </td><td> lim x → c f ( x ) g ( x ) = exp lim x → c g ( x ) 1 / ln f ( x ) {\displaystyle \lim _{x\to c}f(x)^{g(x)}=\exp \lim _{x\to c}{\frac {g(x)}{1/\ln f(x)}}\!} </td><td> lim x → c f ( x ) g ( x ) = exp lim x → c ln f ( x ) 1 / g ( x ) {\displaystyle \lim _{x\to c}f(x)^{g(x)}=\exp \lim _{x\to c}{\frac {\ln f(x)}{1/g(x)}}\!} </td></tr></tbody></table> <h2 id="see-also">See also</h2> <ul><li><a href="/facts/Defined_and_undefined/Bc1Q3px4">Defined and undefined</a></li> <li><a href="/facts/Division_by_zero/wXK7t2Ge">Division by zero</a></li> <li><a href="/facts/Extended_real_number_line/65Gs7QS0">Extended real number line</a></li> <li><a href="/facts/Indeterminate_equation/DwdC3hyK">Indeterminate equation</a></li> <li><a href="/facts/Indeterminate_system/DwdC3hyK">Indeterminate system</a></li> <li><a href="/facts/Indeterminate_(variable)/R7QAHS4j">Indeterminate (variable)</a></li> <li><a href="/facts/L%2527H%25C3%25B4pital%2527s_rule/pCXK75Io">L'Hôpital's rule</a></li></ul> <h3>Citations</h3> <h3>Bibliographies</h3> <ul><li>Varberg, Dale E.; Purcell, Edwin J.; Rigdon, Steven E. (2007). <i>Calculus</i> (9th ed.). <a href="/facts/Pearson_Prentice_Hall/bynWAuat">Pearson Prentice Hall</a>. <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 978-0131469686.</li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>Varberg, Purcell & Rigdon (2007), p. 423, 429, 430, 431, 432. - Varberg, Dale E.; Purcell, Edwin J.; Rigdon, Steven E. (2007). Calculus (9th ed.). Pearson Prentice Hall. ISBN 978-0131469686. <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> <li id="fn:2"><p>Weisstein, Eric W. "Indeterminate". mathworld.wolfram.com. Retrieved 2019-12-02. <a href="http://mathworld.wolfram.com/Indeterminate.html" target="_blank">http://mathworld.wolfram.com/Indeterminate.html</a> <a href="#fnref:2" class="footnote-back-ref">↩</a></p></li> <li id="fn:3"><p>Louis M. Rotando; Henry Korn (January 1977). "The indeterminate form 00". Mathematics Magazine. 50 (1): 41–42. doi:10.2307/2689754. JSTOR 2689754. <a href="/wiki/Doi_(identifier)" target="_blank">/wiki/Doi_(identifier)</a> <a href="#fnref:3" class="footnote-back-ref">↩</a></p></li> <li id="fn:4"><p>"Undefined vs Indeterminate in Mathematics". www.cut-the-knot.org. Retrieved 2019-12-02. <a href="https://www.cut-the-knot.org/blue/GhostCity.shtml" target="_blank">https://www.cut-the-knot.org/blue/GhostCity.shtml</a> <a href="#fnref:4" class="footnote-back-ref">↩</a></p></li> <li id="fn:5"><p>"Table of equivalent infinitesimals" (PDF). Vaxa Software. <a href="http://www.vaxasoftware.com/doc_eduen/mat/infiequi.pdf" target="_blank">http://www.vaxasoftware.com/doc_eduen/mat/infiequi.pdf</a> <a href="#fnref:5" class="footnote-back-ref">↩</a></p></li> </ol>