If we directly evaluate the limit $\lim_{x\to 3}\left(\frac{2x^3-5x^2-2x-3}{4x^3-13x^2+4x-3}\right)$ as $x$ tends to $3$, we can see that it gives us an indeterminate form
$\frac{0}{0}$
2
We can solve this limit by applying L'H么pital's rule, which consists of calculating the derivative of both the numerator and the denominator separately
Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more