Can Graphs Help Estimate Limits in Calculus Problems?

[neutron star]

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

 

[Quincy]

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...