JAMB - Mathematics (2022 - No. 23)
find the limit of y = \(\frac{x^3 + 6x - 7}{x-1}\) as x tends to 1
9
8
0
7
설명
\(\frac{x^3 + 6x - 7}{x-1}\):
When numerator is differentiated → 3x\(^2\) + 6
When denominator is differentiated → 1
: \(\frac{3x^2 + 6}{1}\)
substitute x for 1
\(\frac{3 * 1^2 + 6}{1}\) = \(\frac{3 + 6}{1}\)
= \(\frac{9}{1}\)
= 9