JAMB Mathematics 2022 Paper

Concept:At $x=1$, direct substitution gives the indeterminate form $\frac{0}{0}$, so simplify the expression before finding the limit.Explanation:Substitute $x=1$ into the numerator:$1^3 + 6(1) - 7 = 0$.The denominator also becomes $1-1=0$.Since $x=1$ is a root of $x^3+6x-7$, factorise it:$$x^3+6x-7=(x-1)(x^2+x+7).$$Therefore, for $x\neq 1$,$$\frac{x^3+6x-7}{x-1}=x^2+x+7.$$Now evaluate the limit as $x$ tends to $1$:$$\lim_{x\to 1}(x^2+x+7)=1^2+1+7=9.$$Using L'Hôpital's rule also gives the same result: differentiate the numerator to get $3x^2+6$ and the denominator to get $1$, then substitute $x=1$ to obtain $9$.Answer:$9$, which is option A.