find the limit of y = \(\frac{x^3 + 6x - 7}{x-1}\) as x tends to 1
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Correct Answer: Option A
Explanation:
\(\frac{x^3 + 6x - 7}{x-1}\):
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
\(\frac{x^3 + 6x - 7}{x-1}\):
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