Question

Find n if $\displaystyle \lim_{x \to 2}\dfrac{{{x}^{n}}-{{2}^{n}}}{x-2}=80$ and n being a positive integer.

Answer 1

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.