Question

$\underset{x\to 3}{lim}\frac{x^3-7x^2+15x-9}{x^4-5x^3+27x-27}$ is equal to

• A. $\frac{2}{3}$
• B. $\frac{2}{9}$
• C. $\frac{1}{9}$
• D. $1$

Answer 1

Answer: (B)

$\frac{2}{9}$

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

$=\frac{(3{)}^3-7(3{)}^2+15\left(3\right)-9}{(3{)}^4-5(3^3)+27(3)-27}=\frac{27-63+45-9}{81-135+81-27}=\frac{72-72}{162-162}=\frac{0}{0}$

Apply L - Hospital Rule -1

${lim}_{x\to 3}=\frac{x^3-7x^2+15x-9}{x^4-5x^3+27x-27}={lim}_{x\to 3}\frac{(x^3-7x^2+15x-9{)}^1}{(x^4-5x^3+27x-27{)}^1}$

$={lim}_{x\to 3}\frac{3x^2-14x+15}{4x^3-15x^2+27}$

$=\frac{3(3{)}^2-14(3)+15}{4(3{)}^3-15(3{)}^2+27}=\frac{27-42+15}{108-135+27}=\frac{42-42}{135-135}=\frac{0}{0}$

${lim}_{x\to 3}\frac{x^3-7x^2+15x-9}{x^4-5x^3+27x-27}={lim}_{x\to 3}\frac{(3x^2-14x+15{)}^1}{(4x^3-15x^2+27{)}^1}$

$={lim}_{x\to 3}\frac{6x-14}{12x^2-30x}$

$=\frac{6\left(3\right)-14}{12(3{)}^2-30(3)}=\frac{18-14}{108-90}=\frac{4}{18}=\frac{2}{9}$