Question

The value of $\underset{x\to 3}{lim}\frac{x^3-7x^2+15x-9}{x^4-5x^3+27x-27}$ is $\frac{a}{b}$ in the smallest form. Find $a+b$.

Answer 1

The expression is given as,

$\frac{x^3-7x^2+15x-9}{x^4-5x^3+27x-27}$

Differentiating the numerator and denominator of the above term with respect to $x$, we get

$\frac{3x^2-14x+15}{4x^3-15x^2+27}$

Again differentiating the numerator and denominator of the above term with respect to $x$, we get

$\frac{6x-14}{12x^2-30x}$

Put the limit value in the above term we get

$\frac{a}{b}=\frac{6\times 3-14}{12\times {\left(3\right)}^2-30\times 3}$

$\frac{a}{b}=\frac{2}{9}$

The sum is given as,

$a+b=2+9$

$=11$

Thus the value of $a+b$ is $11$.