Question

Evaluate the following expression: \[{{\lim }_{x\to 2}}\dfrac{{{x}^{5}}-32}{{{x}^{3}}-8}\].

Answer 1

Hint: We will use the knowledge of indeterminate forms and L’Hospital’s rule to solve this problem. First we will put 2 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 answer:
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 \[\dfrac{0}{0}\] 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 \[f(x)\] at \[x\to a\]. Then we first check whether it is an indeterminate form or not by directly putting the value of \[x=a\] in the given function. If we get \[\dfrac{0}{0}\] and \[\dfrac{\infty }{\infty }\] form they are called indeterminate forms. L’Hospital’s Rule is applicable in the two cases.
Now let \[{{\lim }_{x\to 2}}\dfrac{{{x}^{5}}-32}{{{x}^{3}}-8}......(1)\]
So first we will check if this is an indeterminate form or not by putting x as 2 in equation (1) we get,
\[\Rightarrow \dfrac{{{2}^{5}}-32}{{{2}^{3}}-8}=\dfrac{32-32}{8-8}=\dfrac{0}{0}\]. 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,
\[\Rightarrow {{\lim }_{x\to 2}}\dfrac{5{{x}^{4}}}{3{{x}^{2}}}.......(2)\]
Cancelling similar terms in equation (2) we get,
\[\Rightarrow {{\lim }_{x\to 2}}\dfrac{5}{3}{{x}^{2}}........(3)\]
Now substituting the value of x in equation (3) we get,
\[\Rightarrow {{\lim }_{x\to 2}}\dfrac{5}{3}{{x}^{2}}=\dfrac{5}{3}\times {{2}^{2}}=\dfrac{20}{3}\]
Hence the answer is \[\dfrac{20}{3}\].

Note: Remembering 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.