neutron star said:
Ok thanks, how do you solve this type of problem?
http://img156.imageshack.us/img156/7294/picture6rn.png
http://img156.imageshack.us/img156/6981/picture7ki.png
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Ok, suppose k = 4.
So you have: lim as x->infinity (2
4x + 8)/(2
3x + 5). Rather than looking at the function as a whole, look at the numerator and denominator separately.
Consider the dominant term in the function. When x is really large/close to infinity, the value of 2
4x is going to be larger than the value of 2
3x, and definitely larger than the value of 8 and 5. Since 2
4x is the dominant term, only consider the limit for that function. So, the lim as x->infinity 2
4x = ? As x gets larger and larger, the value of 2
4x also gets larger and larger, so the answer is
infinity.If k = 3, then there is no one dominant term. So find the lim as x->infinity (2
3x/2
3x), which is of course equal to lim as x->infinity (
1), since 2
3x/2
3x cancel.
If k< 3, then again, only consider dominant terms. So we have lim as x->infinity (2
2x/2
3x) (suppose k = 2).
As x-> infinity the denominator is going to be becoming much larger than the numerator, so the value of the function is going to get smaller and smaller, and approach
zero.
I know my explanation is fuzzy, I just hope I didn't confuse you...