find the limit of y = \( \frac{x^3 + 6x - 7}{x - 1} \) as x tends to 1

x3+6x−7x−1 $\frac{{\mathrm{<br>}}^{}}{}$

When numerator is differentiated → 3x2 ${}^2$ + 6  

When denominator is differentiated → 1

: 3x2+61 $\frac{\mathrm{<br>}}{}$

substitute x for 1

 3∗12+61 $\frac{3\ast 1^2+6}{1}$ =  3+61 $\frac{3+6}{1}$ 

=  91 $\frac{9}{1}$

= 9