Question

can you check my work please on the precise definition of limit at infinity

Answer 1

there is a mistake in 4th line

Answer 2

^ this guy

Answer 3

\[\begin{align}
&\left|\dfrac{2x-3}{x+2}-2\right|\lt0.01 \\~\\
&\left|\dfrac{2x-3}{x+2} - 2\cdot\dfrac{x+2}{x+2}\right|\lt0.01 \\~\\

\end{align}\]

simplify the fraction inside absolute bars first

Answer 4

you get
\[\left|\dfrac{7}{x+2} \right|\lt0.01\]

Answer 5

M=698 is correct though

Answer 6

oh i see thank you so much :D