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Question: How do you evaluate the limit of \(\lim \dfrac{{{x}^{3}}+8}{{{x}^{2}}-4}\) as \(x \to -2\) ?...

How do you evaluate the limit of limx3+8x24\lim \dfrac{{{x}^{3}}+8}{{{x}^{2}}-4} as x2x \to -2 ?

Explanation

Solution

This limit function can be solved by first simplifying the polynomial. Expand the denominator and write the numerator also in terms of the denominator so that we can easily cancel the terms which would lead to the form 00\dfrac{0}{0} . After eliminating this form, substitute the value of x in the expression and evaluate to get the limit of the function.

Complete step by step solution:
The given limit is limx2x3+8x24\displaystyle \lim_{x \to -2}\dfrac{{{x}^{3}}+8}{{{x}^{2}}-4}
When we substitute the value of x as -2, we can see that it would lead to the form 00\dfrac{0}{0}.
Hence to avoid we shall evaluate or expand the expression and eliminate the terms which would lead to that form.
We expand the denominator first.
limx2x3+8(x+2)(x2)\Rightarrow \displaystyle \lim_{x \to -2}\dfrac{{{x}^{3}}+8}{\left( x+2 \right)\left( x-2 \right)}
Now we shall write the numerator also in terms of the denominator.
The word in terms means that we need to write in such a way such that the common terms in the numerator and denominator get cancelled.
We mainly write the numerator in terms of a particular term in the denominator which leads to a 0. Here when we substitute x as -2 in (x+2)\left( x+2 \right) we get a 0.
limx2(x+2)(x22x+4)(x+2)(x2)\Rightarrow \displaystyle \lim_{x \to -2}\dfrac{\left( x+2 \right)\left( {{x}^{2}}-2x+4 \right)}{\left( x+2 \right)\left( x-2 \right)}
Now cancel the common terms.
After simplification we get,
limx2(x22x+4)(x2)\Rightarrow \displaystyle \lim_{x \to -2}\dfrac{\left( {{x}^{2}}-2x+4 \right)}{\left( x-2 \right)}
We can directly substitute the value of x as -2 since it is no longer in 00\dfrac{0}{0} form.
Upon evaluating we get,
((2)22(2)+4)((2)2)\Rightarrow \dfrac{\left( {{\left( -2 \right)}^{2}}-2\left( -2 \right)+4 \right)}{\left( \left( -2 \right)-2 \right)}
(4+4+4)(4)\Rightarrow \dfrac{\left( 4+4+4 \right)}{\left( -4 \right)}
3\Rightarrow -3
Hence the limit limx2x3+8x24\displaystyle \lim_{x \to -2}\dfrac{{{x}^{3}}+8}{{{x}^{2}}-4} on evaluation gives -3 as the answer.

Note: Whenever a limit is given to evaluate, the simplest way to evaluate is to substitute the value of x. If we get either of the forms 00\dfrac{0}{0} or a0\dfrac{a}{0} then we know that we need to expand the expression by factoring or conjugating or by using the trigonometric identities to eliminate the terms which would lead to the form