What is the limit for x---infinity

(r+sinr) / (2r+7-5sinr)

Question

What is the limit for x--->infinity

(r+sinr) / (2r+7-5sinr)

anonymous · Answer

I guess you mean as r tend to infinity?

Divide the top and bottom of the fraction both by r. Then use the fact:
$$\lim_{r \rightarrow \infty} \frac{\sin r}{r} = 0$$(since sin(x) remains bounded between -1 and 1, whereas r goes off to infinity.)

anonymous · Answer

Yep, woops. Thank you

anonymous · Answer

You're welcome :)

rob1525 · Answer

Um both the top and bottom go to infinity...so...the limit shouldn't exist. let me verify with my calc.

rob1525 · Answer

Nope im getting 1/2. Humm.....

anonymous · Answer

@robtobey Remember in maths $\frac{\infty}{\infty}$ could be anything - it's basically meaningless. Just because the top of the fraction tends to infinity, and the bottom tends to infinity, does NOT mean the limit doesn't exist!

anonymous · Answer

(Sorry, quoted wrong person!)

@rob1525 See above.

rob1525 · Answer

oh, geees. if the sin are eliminated then ur left with x/2x which is a horizontal retricemtote of 1/2.

rob1525 · Answer

*asymptote