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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. 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