Question

lim of x of infinity 3xsquared -x+4 all over 2xsquared + 5x - 8
I have to show each step...help!

Answer 1

There's really no good way to simplify that except to know that when there is a \[\lim_{x \rightarrow \infty}\]then you have to check the powers of the fraction.
In this case, the highest power on top is 3 x squared, and the highest power on bottom is 2 x squared. Since you can cross off the lower powers, you're left with 3x squared over 2x squared which is 3/2.

Answer 2

3/2

Answer 3

what pjw said but you do it with your eyeballs

Answer 4

First : divid all terms of the numerator & denomenator by the highest poer , i.e x^2

Answer 5

however... we can write some steps if you like

Answer 6

\[\frac{3x^2-x+4}{2x^2+5x-8}=\frac{\frac{3x^2}{x^2}-\frac{x}{x^2}+\frac{4}{x^2}}{\frac{2x^2}{x^2}+\frac{5x}{x^2}-\frac{8}{x^2}}\]

Answer 7

you get
\[\frac{3-\frac{1}{x}+\frac{4}{x^2}}{2+\frac{5}{x}-\frac{8}{x^2}}\]

Answer 8

===> lim [( 3+4/x^2) /2+5/x^2-5/x^2] = (3+0)/(2+0+0)=3/2

Answer 9

take the limit as x goes to infinity and see that everything goes to zero except the 3 and the 2 so limit is 3/2
but as you can see this is a colossal waste of time. you do it with your eyes. degree of numerator is the same as the degree of the denominator, take ratio of leading coefficient.

Answer 10

nice! thanx!!!