math

posted by .

i really don't understand why this 1-lim 3/(y+4) [the lim is x->infinity] the ans is 0 while this 2-lim (7-6x^5)/(x+3) [x->+infinity ] is - infinity.
can anyone explain why the 1 is zero & the 2 is infinity.

• math -

Try to get a "feel" for numbers.

look at 3/(y+4)
As the y+4 gets larger and larger, and since the top stays constant at 3, the quotient gets smaller and smaller.
e.g. let y = 1 000 000, (not even close to infinity)
the quotient is 3/(1000000 +4) = 0.000003
pretty close to zero, but we have a long way to go in making y ---> infinity

for the (7 - 6x^5)/(x+3) , x--->+infinity

pick a "large value of x" , say x = 1000

so you would get (7 - 6000 000 000 000 000)/1003
would you not agree that this is a huge negative number ? And our x is not even remotely "large".

You might even want to experiment with your calculator.

Similar Questions

1. calculus - interval of convergence

infinity of the summation n=0: ((n+2)/(10^n))*((x-5)^n) .. my work so far. i used the ratio test = lim (n-->infinity) | [((n+3)/(10^(n+1)))*((x-5)^(n+1))] / [((n+2)/(10^n))*((x-5)^n)] | .. now my question is: was it ok for me to …
2. calculus - interval of convergence

infinity of the summation n=0: ((n+2)/(10^n))*((x-5)^n) .. my work so far. i used the ratio test = lim (n-->infinity) | [((n+3)/(10^(n+1)))*((x-5)^(n+1))] / [((n+2)/(10^n))*((x-5)^n)] | .. now my question is: was it ok for me to …
3. Pre-cal

Please determine the following limits if they exist. If the limit does not exist put DNE. lim 2+6x-3x^2 / (2x+1)^2 x-> - infinity lim 4n-3 / 3n^2+2 n-> infinity I did lim 2+6x-3x^2 / (2x+1)^2 x-> - infinity (2+6x-3x²)/(4x²+4x+1) …
4. Calc. Limits

Are these correct? lim x->0 (x)/(sqrt(x^2+4) - 2) I get 4/0= +/- infinity so lim x->0+ = + infinity?
5. calc

Are these correct? lim x->0 (x)/(sqrt(x^2+4) - 2) I get 4/0= +/- infinity so lim x->0+ = + infinity?

Are these correct? lim x->0 (x)/(sqrt(x^2+4) - 2) I get 4/0= +/- infinity so lim x->0+ = + infinity?
7. Calculus

Find the horizontal asymptote of f(x)=e^x - x lim x->infinity (e^x)-x= infinity when it's going towards infinity, shouldn't it equal to negative infinity, since 0-infinity = - infinity lim x-> -infinity (e^x)-x= infinity
8. calculus

State which of the conditions are applicable to the graph of y = f(x). (Select all that apply.) lim x→infinity f(x) = −infinity lim x→a+ f(x) = L lim x→infinity f(x) = L f is continuous on [0, a] lim x→infinity …
9. Math

1. If -1/infinity = infinity or -infinity ?
10. Math

1. If -1/infinity = infinity or -infinity ?

More Similar Questions