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Evaluate: lim\(_{x→-2}\) \(\frac{x^3+8}{x+2}\).

Further Mathematics
WAEC 2022

Evaluate: lim\(_{x→-2}\) \(\frac{x^3+8}{x+2}\).

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Explanation

\(\frac{x^3+8}{x+2}\).

x\(^3\) + 8 = x\(^3\) + 23
recall that
x\(^3\) + y\(^3\) = (x+y)(x\(^2\) - xy + y\(^2\))
x = x, y = 2
x\(^3\) + 8 = (x+2)(x\(^3\) - 2(x) + 22)

\(\frac{x^3+8}{x+2} = \frac{(x+2)(x^3 - 2(x) + 22)}{x+2}\) → x\(^2\) - 2x + 4

x = -2
(-2)\(^2\) - 2(2) + 4 = 4 - 4 + 4

= 4

 


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WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
WAEC offline past questions - with all answers and explanations in one app - Download for free
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
WAEC offline past questions - with all answers and explanations in one app - Download for free
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709