Question

If $\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^n} - {2^n}}}{{x - 2}} = 448$, then $n =$

Answer 1

First, we will see the concept of limits.
Limits are defined as a function that has some value that approaches the input. Function with the limit is represented as $\mathop {\lim }\limits_{x \to a } f(x) = L$.
If the limits values get zero or infinity in both forms $\dfrac{0}{0},\dfrac{\infty }{\infty }$all we need to do is to differentiate the numerator and denominator for the general form and then take the limit.
Suppose after the differentiation also if the terms are zero or infinity then do the same process again and again until you reach an existing point. (L’Hospital rule)

Complete step by step answer:
From the given that $\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^n} - {2^n}}}{{x - 2}} = 448$ and we need to find the value of n.
Now applying the limit value with $x \to 2$(from the given), to the limit point $\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^n} - {2^n}}}{{x - 2}} = 448$, we get after applying x as the number two$\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^n} - {2^n}}}{{x - 2}} = 448 \Rightarrow \dfrac{{{2^n} - {2^n}}}{{2 - 2}} = 448$.
Further solving this we get, $\dfrac{{{2^n} - {2^n}}}{{2 - 2}} = 448 \Rightarrow \dfrac{0}{0} = 448$(which is the form of zero by zero)
So, we don’t get any value from the limits, thus by the use of L hospital use we can now differentiate both terms as follows, $\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^n} - {2^n}}}{{x - 2}} = 448$ only differentiate the limit values from the left-hand side (in the right side there is no limit function)
Thus, we get, $\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^n} - {2^n}}}{{x - 2}} = 448 \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{\dfrac{d}{{dx}}({x^n} - {2^n})}}{{\dfrac{d}{{dx}}(x - 2)}} = 448$(differentiating with respect to x)
Hence in the denominator, the value gets one because $\dfrac{d}{{dx}}(x - 2) = 1 - 0 \Rightarrow 1$ and in the numerator we get, $\dfrac{d}{{dx}}({x^n} - {2^n}) = n{x^{n - 1}} - 0 \Rightarrow n{x^{n - 1}}$ (${2^n}$is don’t have any values in x; thus, it turns to zero)
Hence, we get, $\mathop {\lim }\limits_{x \to 2} \dfrac{{\dfrac{d}{{dx}}({x^n} - {2^n})}}{{\dfrac{d}{{dx}}(x - 2)}} = 448 \Rightarrow \mathop {\lim }\limits_{x \to 2} n{x^{n - 1}} = 448$, now apply the limit as $x \to 2$and we get, $\mathop {\lim }\limits_{x \to 2} n{x^{n - 1}} = 448 \Rightarrow n{2^{n - 1}} = 448$
Now we will convert the number $448$into the form of $7 \times 64$(we can able to compare this in left side values and try to cancel each other)
Thus, we get, $448 = 7 \times {2^{7 - 1}}$($64 = {2^6} \Rightarrow {2^{7 - 1}}$rewritten to cancel common terms)
Hence, we get, $n{2^{n - 1}} = 448 \Rightarrow n{2^{n - 1}} = 7 \times {2^{7 - 1}}$, as we see that both the value is common but the only difference is in left-hand side we have n, and in the right side, we have seven.
Since both the values are in $=$terms.
Thus, we get, the value of n is $n = 7$
Therefore, $\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^n} - {2^n}}}{{x - 2}} = 448$ where $n = 7$

Note: From the definition of the limits, take two terms like $p(x),q(x)$then some formulas are using this in the limits.
$\mathop {\lim }\limits_{x \to \infty } [p(x) + q(x)] = \mathop {\lim }\limits_{x \to \infty } p(x) + \mathop {\lim }\limits_{x \to \infty } q(x)$
$\mathop {\lim }\limits_{x \to \infty } \dfrac{{p(x)}}{{q(x)}} = \dfrac{{\mathop {\lim }\limits_{x \to \infty } p(x)}}{{\mathop {\lim }\limits_{x \to \infty } q(x)}}$
$\mathop {\lim }\limits_{x \to \infty } [p(x)q(x)] = \mathop {\lim }\limits_{x \to \infty } p(x) \times \mathop {\lim }\limits_{x \to \infty } q(x)$