Asked by susan
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.
can anyone explain why the 1 is zero & the 2 is infinity.
Answers
Answered by
Reiny
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.
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.
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