the limit of ((sqrt(5x^4+2x))/x^2) as x approaches infinity?
i know the answer is sqrt 5, but i cant seem to remember how to solve it.
lol, im reviewing for midterms which i have next week, and i totally forgot how to do limits which are like the easiest part of calculus...lol!
i actually find derivatives much easier :)
any help greatly appreciated.
thanks!
2007-01-16 09:29:32 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
yea but u can actually take the x^4 out of the sqrt sign?
2007-01-16 09:45:39 · update #1
When x aproaches infinity, you can neglect all lower order terms. Therefore,
5x^4+2x becomes 5x^4
√(5x^4)/x^2 = √5
2007-01-16 09:39:41 · answer #1 · answered by sahsjing 7 · 1⤊ 0⤋
It's because as x gets really big the 5x^4 term overshadows the 2x in the numerator, basically making it negligible. So you have sqrt(5x^4) over x^2, which is sqrt(5) times x^2 over x^2, which is sqrt 5.
2007-01-16 17:38:04 · answer #2 · answered by hayharbr 7 · 1⤊ 0⤋
Here is how it goes:
take the x^2 and put it into the sqr ie sqr[5x^4/x^4 + 2x/x4], so the limit as x-->infinity() = lim sqr[5+2/x^3] = sqr[5] as the 1/x^3 --> 0
2007-01-16 17:36:47 · answer #3 · answered by kellenraid 6 · 0⤊ 0⤋
Solve it using logic
SQRT(5*X^4 + 2X)/X^2-->For large X
2X is much less than X^4-->
SQRT(5*X^4)/X^2-->
SQRT(5)*(X*X)/(X*X)-->
SQRT(5)
2007-01-16 17:41:52 · answer #4 · answered by John W 3 · 1⤊ 0⤋