Find N If Displaystyle Limx To 2dfracxn2nx280 And N Class 11 Maths ...
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Find n if \[\displaystyle \lim_{x \to 2}\dfrac{{{x}^{n}}-{{2}^{n}}}{x-2}=80\] and n being a positive integer.Answer
Verified558k+ viewsHint: In this question, we need to find the value of n if \[\displaystyle \lim_{x \to 2}\dfrac{{{x}^{n}}-{{2}^{n}}}{x-2}=80.\] Put x = 2, and will give us \[\dfrac{0}{0}\] form. So, we will first use L’Hopital’s rule and then evaluate the limit. After evaluating the limit, it will be in the form of x so we will equate it to 80 and get the value of n. According to L’Hopital’s rule, if \[\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}\] is in \[\left( \dfrac{0}{0} \right)\] form, then we can take \[\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}=\displaystyle \lim_{x \to a}\dfrac{{{f}^{'}}\left( x \right)}{{{g}^{'}}\left( x \right)}.\]Complete step-by-step answer:Here, we are given that the function of limit as \[\displaystyle \lim_{x \to 2}\dfrac{{{x}^{n}}-{{2}^{n}}}{x-2}.\] Let us try to evaluate the limit. Putting x = 2, it will give us \[\left( \dfrac{0}{0} \right)\] form, so it is in indeterminate form. Hence, we need to apply L’Hopital’s rule according to which \[\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}=\displaystyle \lim_{x \to a}\dfrac{{{f}^{'}}\left( x \right)}{{{g}^{'}}\left( x \right)}.\] If \[\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}\] is in indeterminate form \[\left( \dfrac{0}{0}\text{or}\dfrac{\infty }{\infty } \right).\] Hence, we need to take the derivative of \[{{x}^{n}}-{{2}^{n}}\] in numerator and the derivative of x – 2 in the denominator. We know that \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\] and \[\dfrac{d}{dx}\left( a \right)=0\] where ‘a’ is constant. So, the derivative of \[{{x}^{n}}-{{2}^{n}}=n{{x}^{n-1}}-0=n{{x}^{n-1}}\] (because \[{{2}^{n}}\] is constant). We know that \[\dfrac{d}{dx}\left( x \right)=1\] and \[\dfrac{d}{dx}\left( 2 \right)=0.\] So, the derivative of x – 2 = 1 – 0 = 1. Now, \[\displaystyle \lim_{x \to 2}\dfrac{{{x}^{n}}-{{2}^{n}}}{x-2}=\displaystyle \lim_{x \to 2}\dfrac{\dfrac{d}{dx}\left( {{x}^{n}}-{{2}^{n}} \right)}{\dfrac{d}{dx}\left( x-2 \right)}\]So calculated, the derivative of \[{{x}^{n}}-{{2}^{n}}\] is \[n{{x}^{n-1}}\] and the derivative of x – 2 is 1. Hence, we get,\[\displaystyle \lim_{x \to 2}\dfrac{{{x}^{n}}-{{2}^{n}}}{x-2}=\displaystyle \lim_{x \to 2}\dfrac{n{{x}^{n-1}}}{1}\]\[\displaystyle \lim_{x \to 2}\dfrac{{{x}^{n}}-{{2}^{n}}}{x-2}=\displaystyle \lim_{x \to 2}n{{x}^{n-1}}\]Now, evaluating the limit on the right side, by putting x = 2, we get, \[\displaystyle \lim_{x \to 2}\dfrac{{{x}^{n}}-{{2}^{n}}}{x-2}=n{{2}^{n-1}}\]But we are given \[\displaystyle \lim_{x \to 2}\dfrac{{{x}^{n}}-{{2}^{n}}}{x-2}\]to be equal to 80. So, we can say that \[{{n}^{2n-1}}\] is equal to 80. So, \[{{n}^{2n-1}}=80\]As we know that 80 can be written as \[5\times 16,\] so,\[n{{2}^{n-1}}=5\times 16\]Also, we know that, \[{{2}^{4}}=16.\]So, we get,\[n{{2}^{n-1}}=5\times {{2}^{4}}\]Now we can write 4 as 5 – 1, so we get, \[\Rightarrow n{{2}^{n-1}}=5\times {{2}^{5-1}}\]By comparing, we get n = 5. Here, n = 5 is the required answer. Note: Students should note that indeterminate form has types \[\dfrac{0}{0},\dfrac{\infty }{\infty },0\times \infty ,\infty -\infty .\] If our limit is any of these forms then we can apply L’Hopital’s. For \[\dfrac{0}{0},\dfrac{\infty }{\infty },\] L’Hopital’s rule is applied directly but for \[0\times \infty ,\infty -\infty \]we have to first convert them into \[\dfrac{0}{0}\text{or}\dfrac{\infty }{\infty }\] form. While comparing \[n{{2}^{n-1}}=80,\] we can just use the trial and error method. Recently Updated PagesMaster Class 12 Economics: Engaging Questions & Answers for Success
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AnswerQuestion Answers for Class 12Class 12 BiologyClass 12 ChemistryClass 12 EnglishClass 12 MathsClass 12 PhysicsClass 12 Social ScienceClass 12 Business StudiesClass 12 EconomicsQuestion Answers for Class 11Class 11 EconomicsClass 11 Computer ScienceClass 11 BiologyClass 11 ChemistryClass 11 EnglishClass 11 MathsClass 11 PhysicsClass 11 Social ScienceClass 11 AccountancyClass 11 Business StudiesQuestion Answers for Class 10Class 10 ScienceClass 10 EnglishClass 10 MathsClass 10 Social ScienceClass 10 General KnowledgeQuestion Answers for Class 9Class 9 General KnowledgeClass 9 ScienceClass 9 EnglishClass 9 MathsClass 9 Social ScienceQuestion Answers for Class 8Class 8 ScienceClass 8 EnglishClass 8 MathsClass 8 Social ScienceQuestion Answers for Class 7Class 7 ScienceClass 7 EnglishClass 7 MathsClass 7 Social ScienceQuestion Answers for Class 6Class 6 ScienceClass 6 EnglishClass 6 MathsClass 6 Social ScienceQuestion Answers for Class 5Class 5 ScienceClass 5 EnglishClass 5 MathsClass 5 Social ScienceQuestion Answers for Class 4Class 4 ScienceClass 4 EnglishClass 4 MathsFind n if \[\displaystyle \lim_{x \to 2}\dfrac{{{x}^{n}}-{{2}^{n}}}{x-2}=80\] and n being a positive integer.AnswerVerified558k+ viewsHint: In this question, we need to find the value of n if \[\displaystyle \lim_{x \to 2}\dfrac{{{x}^{n}}-{{2}^{n}}}{x-2}=80.\] Put x = 2, and will give us \[\dfrac{0}{0}\] form. So, we will first use L’Hopital’s rule and then evaluate the limit. After evaluating the limit, it will be in the form of x so we will equate it to 80 and get the value of n. According to L’Hopital’s rule, if \[\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}\] is in \[\left( \dfrac{0}{0} \right)\] form, then we can take \[\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}=\displaystyle \lim_{x \to a}\dfrac{{{f}^{'}}\left( x \right)}{{{g}^{'}}\left( x \right)}.\]Complete step-by-step answer:Here, we are given that the function of limit as \[\displaystyle \lim_{x \to 2}\dfrac{{{x}^{n}}-{{2}^{n}}}{x-2}.\] Let us try to evaluate the limit. Putting x = 2, it will give us \[\left( \dfrac{0}{0} \right)\] form, so it is in indeterminate form. Hence, we need to apply L’Hopital’s rule according to which \[\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}=\displaystyle \lim_{x \to a}\dfrac{{{f}^{'}}\left( x \right)}{{{g}^{'}}\left( x \right)}.\] If \[\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}\] is in indeterminate form \[\left( \dfrac{0}{0}\text{or}\dfrac{\infty }{\infty } \right).\] Hence, we need to take the derivative of \[{{x}^{n}}-{{2}^{n}}\] in numerator and the derivative of x – 2 in the denominator. We know that \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\] and \[\dfrac{d}{dx}\left( a \right)=0\] where ‘a’ is constant. So, the derivative of \[{{x}^{n}}-{{2}^{n}}=n{{x}^{n-1}}-0=n{{x}^{n-1}}\] (because \[{{2}^{n}}\] is constant). We know that \[\dfrac{d}{dx}\left( x \right)=1\] and \[\dfrac{d}{dx}\left( 2 \right)=0.\] So, the derivative of x – 2 = 1 – 0 = 1. Now, \[\displaystyle \lim_{x \to 2}\dfrac{{{x}^{n}}-{{2}^{n}}}{x-2}=\displaystyle \lim_{x \to 2}\dfrac{\dfrac{d}{dx}\left( {{x}^{n}}-{{2}^{n}} \right)}{\dfrac{d}{dx}\left( x-2 \right)}\]So calculated, the derivative of \[{{x}^{n}}-{{2}^{n}}\] is \[n{{x}^{n-1}}\] and the derivative of x – 2 is 1. Hence, we get,\[\displaystyle \lim_{x \to 2}\dfrac{{{x}^{n}}-{{2}^{n}}}{x-2}=\displaystyle \lim_{x \to 2}\dfrac{n{{x}^{n-1}}}{1}\]\[\displaystyle \lim_{x \to 2}\dfrac{{{x}^{n}}-{{2}^{n}}}{x-2}=\displaystyle \lim_{x \to 2}n{{x}^{n-1}}\]Now, evaluating the limit on the right side, by putting x = 2, we get, \[\displaystyle \lim_{x \to 2}\dfrac{{{x}^{n}}-{{2}^{n}}}{x-2}=n{{2}^{n-1}}\]But we are given \[\displaystyle \lim_{x \to 2}\dfrac{{{x}^{n}}-{{2}^{n}}}{x-2}\]to be equal to 80. So, we can say that \[{{n}^{2n-1}}\] is equal to 80. So, \[{{n}^{2n-1}}=80\]As we know that 80 can be written as \[5\times 16,\] so,\[n{{2}^{n-1}}=5\times 16\]Also, we know that, \[{{2}^{4}}=16.\]So, we get,\[n{{2}^{n-1}}=5\times {{2}^{4}}\]Now we can write 4 as 5 – 1, so we get, \[\Rightarrow n{{2}^{n-1}}=5\times {{2}^{5-1}}\]By comparing, we get n = 5. Here, n = 5 is the required answer. Note: Students should note that indeterminate form has types \[\dfrac{0}{0},\dfrac{\infty }{\infty },0\times \infty ,\infty -\infty .\] If our limit is any of these forms then we can apply L’Hopital’s. For \[\dfrac{0}{0},\dfrac{\infty }{\infty },\] L’Hopital’s rule is applied directly but for \[0\times \infty ,\infty -\infty \]we have to first convert them into \[\dfrac{0}{0}\text{or}\dfrac{\infty }{\infty }\] form. While comparing \[n{{2}^{n-1}}=80,\] we can just use the trial and error method. Recently Updated PagesMaster Class 12 Economics: Engaging Questions & Answers for SuccessMaster Class 12 Maths: Engaging Questions & Answers for SuccessMaster Class 12 Biology: Engaging Questions & Answers for SuccessMaster Class 12 Physics: Engaging Questions & Answers for SuccessMaster Class 8 Maths: Engaging Questions & Answers for SuccessClass 8 Question and Answer - Your Ultimate Solutions GuideMaster Class 12 Economics: Engaging Questions & Answers for SuccessMaster Class 12 Maths: Engaging Questions & Answers for SuccessMaster Class 12 Biology: Engaging Questions & Answers for SuccessMaster Class 12 Physics: Engaging Questions & Answers for SuccessMaster Class 8 Maths: Engaging Questions & Answers for SuccessClass 8 Question and Answer - Your Ultimate Solutions Guide- 1
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