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Question: Evaluate \(\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^{10}} - 1024}}{{{x^5} - 32}}\)....

Evaluate lim⁡x→2x10−1024x5−32\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^{10}} - 1024}}{{{x^5} - 32}}.

Explanation

Solution

Direct substitution is the easiest and pronounced method of evaluating the limits of the function, but this method has a limitation for the indeterminate form of function. To overcome the issue, L’Hospital rule is used to convert the indeterminate functions to the defined function by differentiating the numerator and the denominator separately with respect to the available parameter and then substitute the variable’s value to get the result.

Here, in the question, we need to check that the function is returning a defined constant term or not or direct substitution method, and if not, then we will use the L’Hospital rule here.

Complete step by step answer:
Substitute x=2x = 2 in the function f(x)=x10−1024x5−32f(x) = \dfrac{{{x^{10}} - 1024}}{{{x^5} - 32}} to check whether the given function is giving a defined constant value or not. If the function returns an indeterminate form then, we have to further use the L’Hospital rule to carry out our answer.

At, x=2x = 2 the value of the function is:
f(x=2)=x10−1024x5−32 =(2)10−1024(2)5−32 =1024−102432−32 =00  f(x = 2) = \dfrac{{{x^{10}} - 1024}}{{{x^5} - 32}} \\\ = \dfrac{{{{(2)}^{10}} - 1024}}{{{{(2)}^5} - 32}} \\\ = \dfrac{{1024 - 1024}}{{32 - 32}} \\\ = \dfrac{0}{0} \\\

As, 00\dfrac{0}{0} is an indeterminate form so, we have to apply the L’hospital rule in the given function as:
⇒L(x10−1024x5−32)=(ddx(x10−1024))(ddx(x5−32)) =10x95x4 =2x5  \Rightarrow L\left( {\dfrac{{{x^{10}} - 1024}}{{{x^5} - 32}}} \right) = \dfrac{{\left( {\dfrac{d}{{dx}}\left( {{x^{10}} - 1024} \right)} \right)}}{{\left( {\dfrac{d}{{dx}}\left( {{x^5} - 32} \right)} \right)}} \\\ = \dfrac{{10{x^9}}}{{5{x^4}}} \\\ = 2{x^5} \\\

Now, substitute x=2x = 2 in the function 2x52{x^5} to check whether the given function is giving a defined constant value or not.
⇒\Rightarrow 2x5=2(2)5=642{x^5} = 2{(2)^5} = 64

Hence, the value of the lim⁡x→2x10−1024x5−32\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^{10}} - 1024}}{{{x^5} - 32}} is 64.

Note: Students should be aware while applying the limits directly to the given function, if the constant value is not coming even after using the L’Hospital rule one time, then the same method has to be used until we get a constant answer.