Find the limit of y frac x 3 6x - 7 x-1 as x te... - JAMB Mathematics 2022 Question
Find the limit of y = \(\frac{x^3 + 6x - 7}{x-1}\) as x tends to 1
A
9
B
8
C
0
D
7
correct option: a
\(\frac{x^3 + 6x - 7}{x-1}\):
When numerator is differentiated => 3x\(^2\) + 6
When the denominator is differentiated => 1
Hence,
\(\frac{3x^2 + 6}{1}\)
Substitute x for 1
\(\frac{3 * 1^2 + 6}{1}\) = \(\frac{3 + 6}{1}\)
= \(\frac{9}{1}\)
= 9
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