- #1
Ertosthnes
- 49
- 0
Consider f(x) = x + sin(x)cos(x) and g(x) = e[tex]^{sin(x)}[/tex](x + sin(x)cos(x))
Prove that lim[tex]_{x\rightarrow\infty}[/tex] f(x)/g(x) does not exist but that lim[tex]_{x\rightarrow\infty}[/tex] f '(x)/g '(x) = 0. Explain this.
---------------------
Well, it's pretty straightforward to see that lim f(x)/g(x) does not exist, since you end up with lim e[tex]^{-sin(x)}[/tex].
But for the life of me I can't seem to work out how lim f '(x)/g '(x) = 0.
According to L'Hospital's rule, a situation like this should never occur, but I think that it is also important to note that the equations do not meet the condition in L'Hospital's rule:
If f(a) = 0 and g(a) = 0, then lim[tex]_{x\rightarrow\alpha}[/tex] f(x)/g(x) = lim[tex]_{x\rightarrow\alpha}[/tex] f '(x)/g '(x), provided that the limit on the right exists.
In the problem, the limit on the right exists, but the limit on the left is not equal to the one on the right. Also, f(x) and g(x) do not equal zero when x is infinity.
Can anyone help?
Prove that lim[tex]_{x\rightarrow\infty}[/tex] f(x)/g(x) does not exist but that lim[tex]_{x\rightarrow\infty}[/tex] f '(x)/g '(x) = 0. Explain this.
---------------------
Well, it's pretty straightforward to see that lim f(x)/g(x) does not exist, since you end up with lim e[tex]^{-sin(x)}[/tex].
But for the life of me I can't seem to work out how lim f '(x)/g '(x) = 0.
According to L'Hospital's rule, a situation like this should never occur, but I think that it is also important to note that the equations do not meet the condition in L'Hospital's rule:
If f(a) = 0 and g(a) = 0, then lim[tex]_{x\rightarrow\alpha}[/tex] f(x)/g(x) = lim[tex]_{x\rightarrow\alpha}[/tex] f '(x)/g '(x), provided that the limit on the right exists.
In the problem, the limit on the right exists, but the limit on the left is not equal to the one on the right. Also, f(x) and g(x) do not equal zero when x is infinity.
Can anyone help?
Last edited: