Question

lim->∞ (√(t)+t^2)/(3t-t^2) anyone know how to solve this?

Answer 1

is it
\[\lim_{t\to \infty}\frac{\sqrt{t+t^2}}{3-t^2}\]?

Answer 2

if so the limit is zero, because the denominator has a \(t^2\) which grows faster than \(\sqrt{t^2}\) as the second one grows like \(t\)

Answer 3

no there is only one squareroot which is over the t so it is √t + t^2

Answer 4

answer is -1 but idk how to get that

Answer 5

oooho in that case it is \(-1\)

Answer 6

can you show the work? cause thats what im trying to figure out

Answer 7

\[\lim_{t\to \infty}\frac{\sqrt{t}+t^2}{3t-t^2}\] ignore everything but the highest powers, which is 2
it is the ratio of the leading coefficients which is \(-1\)

Answer 8

your teacher probably wants you to divide everything top and bottom by \(t^2\) to get
\[\frac{\frac{1}{\sqrt{t}}+1}{\frac{3}{t}-1}\] and then as \(t\to \infty\) the first term in the numerator and denominator go to zero, leaving \(\frac{1}{-1}=-1\)

Answer 9

actually you get
\[\frac{\frac{1}{\sqrt{t^3}}+1}{\frac{3}{t}-1}\] but same idea

Answer 10

how did the √t^3 get to the bottom after dividing everything by t^2?