Evaluate the limit without using L' Hospital Rule.

Question

Answer 1

Let $f(x) = x^3-7x^2+15x-9; \ g(x) = x^4-5x^3+27x-27$

Using hit and trial $f(1) = 0$, hence $f(x) = (x-1)(x^2-6x+9) = (x-1)(x-3)^2$

We got $1,3,3$ as roots of $f(x)$, so lets try them to find if any of those are roots of $g(x)$ too.

$g(1) = -4 \neq 0$ (not a root)

$g(3) = 0$ (its a root)

$g(x) = (x-3)(\underbrace{x^3-2x^2-6x+9}_{G(x)})$

Now, $G(3) = 0$ (its a root) $\implies G(x) = (x-3)(x^2+x-3)$

Hence, $g(x) = (x-3)^2(x^2+x-3)$

$$\begin{align}
\lim_{x \to 3} \ \frac{f(x)}{g(x)} &= \lim_{x \to 3} \ \frac{(x-1)\cancel{(x-3)^2}}{\cancel{(x-3)^2}(x^2+x-3)} \\
&= \lim_{x \to 3} \ \frac{x-1}{x^2+x-3} \\
&= \frac{3-1}{9+3-3} \\
&= \frac{2}{9}
\end{align}$$

$\textbf{Option (A) is correct}$

Answer 2

solve by this approach