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Question: What is \(\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^5} - 32}}{{x - 2}}?\)...

What is limx2x532x2?\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^5} - 32}}{{x - 2}}?

Explanation

Solution

We will use the knowledge of indeterminate forms and L’Hospital’s rule to solve this problem. First we will put 22 in place of x in the expression and check if it is coming in indeterminate form or not. If yes then we will differentiate both numerator and denominator until the indeterminate form disappears.

Complete step-by-step solution:
Before proceeding with the question we should understand the concept of L’Hospital’s rule for solving indeterminate forms.
In calculus, L’Hospital’s rule is a powerful tool to evaluate limits of indeterminate forms. This rule will be able to show that a limit exists or not, if yes then we can determine its exact value. In short, this rule tells us that in case we are having indeterminate forms like 00\dfrac{0}{0} and \dfrac{\infty }{\infty } then we just differentiate the numerator as well as the denominator and simplify evaluation of limits.
Suppose we have to calculate a limit of f(x)f(x) at xax \to a . Then we first check whether it is an indeterminate form or not by directly putting the value of x=ax = a in the given function. If you get 00\dfrac{0}{0} and \dfrac{\infty }{\infty }form they are called indeterminate forms. L’Hospital’s Rule is applicable in the two cases.
Now let limx2x532x2\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^5} - 32}}{{x - 2}} …………………………(1)
So first we will check if this is an indeterminate form or not by putting x as 22 in equation (1) , we get
x532x2=323222\dfrac{{{x^5} - 32}}{{x - 2}} = \dfrac{{32 - 32}}{{2 - 2}}
=00= \dfrac{0}{0}
So yes this is an indeterminate form and now we will apply L’Hospital’s rule in equation (1) by differentiating both numerator and denominator and hence we get,
limx25x41\Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{5{x^4}}}{1} ………………………………..(2)
Now substituting the value of x in equation (2) , we get
limx25x41=5×241\Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{5{x^4}}}{1} = \dfrac{{5 \times {2^4}}}{1}
=5×16= 5 \times 16
=80= 80
Hence the answer is 8080 .

Note: Remember about L’Hospital’s rule and indeterminate forms is the key here. Also differentiation of x to the power something should be known and we have to keep in mind that differentiation of a constant is always zero. We can make a mistake in differentiating equation (1) so we need to be careful while doing this step.