Question

Find the limit, if it exists. If it does not exist, enter DNE.
lim x^3 − 2x + 3/
x-> ∞ 4 − 5x^2
I answered ∞, but it was wrong.
Please explain how to get to the answer, thank you.

Answer 1

\(\large \begin{align} \color{black}{
\lim_{x\to\infty} \dfrac{x^3-2x+3}{4-5x^2}\hspace{.33em}\\~\\
\normalsize \text{applying L'hopital's rule}\hspace{.33em}\\~\\
=\lim_{x\to\infty} \dfrac{\dfrac{d}{dx}\left(x^3-2x+3\right)}{\dfrac{d}{dx}\left(4-5x^2\right)}\hspace{.33em}\\~\\
=\lim_{x\to\infty} \dfrac{\left(3x^2-2\right)}{\left(-10x\right)}\hspace{.33em}\\~\\
\normalsize \text{applying L'hopital's rule again} \hspace{.33em}\\~\\
=\lim_{x\to\infty} \dfrac{\dfrac{d}{dx}\left(3x^2-2\right)}{\dfrac{d}{dx}\left(-10x\right)}\hspace{.33em}\\~\\
=\lim_{x\to\infty} \dfrac{6x}{-10}\hspace{.33em}\\~\\
=-\dfrac{3\times \infty }{5}\hspace{.33em}\\~\\
=-\infty
}\end{align}\)

Answer 2

but m not sure if this solution is totally correct