Question

Find the value of K that would make the limit exist. Find the limit.
lim(x->0) (2x^3-6)/(x^k+3)

Answer 1

I meant approaches inifinity

Answer 2

\(\large \begin{align} \color{black}{ \lim_{x\to\infty}\frac{2x^3-6}{x^k+3} \hspace{.33em}\\~\\
\normalsize \text{if i put k=3 then } \hspace{.33em}\\~\\
= \lim_{x\to\infty}\frac{2x^3-6}{x^3+3} \hspace{.33em}\\~\\
= \lim_{x\to\infty}\frac{2-\frac{6}{x^3}}{1+\frac{3}{x^3}} \hspace{.33em}\\~\\
=\frac{2-\frac{6}{\infty^3}}{1+\frac{3}{\infty^3}} \hspace{.33em}\\~\\
=\frac{2-0}{1+0} \hspace{.33em}\\~\\
=2 \hspace{.33em}\\~\\
\normalsize \text{so for k=3 the limit exists } \hspace{.33em}\\~\\
}\end{align}\)

Answer 3

Thank you very much, I understand it now.

Answer 4

yw

Answer 5

In fact, any value of \(k\ge3\) would work.