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
Verified567.6k+ 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 11 Business Studies: 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.AnswerVerified567.6k+ 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 11 Business Studies: Engaging Questions & Answers for SuccessMaster Class 11 Computer Science: Engaging Questions & Answers for SuccessMaster Class 11 Maths: Engaging Questions & Answers for SuccessMaster Class 11 Chemistry: Engaging Questions & Answers for SuccessWhich cell organelles are present in white blood C class 11 biology CBSEWhat is the molecular geometry of BrF4 A square planar class 11 chemistry CBSEMaster Class 11 Business Studies: Engaging Questions & Answers for SuccessMaster Class 11 Computer Science: Engaging Questions & Answers for SuccessMaster Class 11 Maths: Engaging Questions & Answers for SuccessMaster Class 11 Chemistry: Engaging Questions & Answers for SuccessWhich cell organelles are present in white blood C class 11 biology CBSEWhat is the molecular geometry of BrF4 A square planar class 11 chemistry CBSE- 1
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