find the limit of y = \(\frac{x^3 + 6x – 7}{x-1}\) as x tends to 1
-
9
- 8
- 0
- 7
The correct answer is: A
Explanation
\(\frac{x^3 + 6x - 7}{x-1}\):
When numerator is differentiated β 3x\(^2\) + 6
When denominator is differentiated β 1
: \(\frac{3x^2 + 6}{1}\)
substitute x for 1
\(\frac{3 * 1^2 + 6}{1}\) = \(\frac{3 + 6}{1}\)
= \(\frac{9}{1}\)
= 9
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