Without using L’Hôpital’s Rule, evaluate the following limit. Use ∞,−∞ to denote divergence to +∞,−∞ and DNE to indicate an other type of non-convergence.

Question

baldymcgee6 · Answer

$$\lim_{x \rightarrow \infty} (2\sin(lnx) +1)/x^2$$

anonymous · Answer

Funny really .... the denumerator is bounded.... and the denumerator goes to infinity.
Any questions ?!

anonymous · Answer

numerator is bounded, denominator is unbounded.... so limit is zero?

anonymous · Answer

Are you the 99% that own the 1% OR the 1% that owns the 99 ?

baldymcgee6 · Answer

@Mikael I have no idea what you are talking about

anonymous · Answer

basics of limits$$\frac{ bounded-function }{ \infty } = 0$$

baldymcgee6 · Answer

So how do I do my problem? haha

anonymous · Answer

sine is stuck between -1 and 1, so numerator is stuck between -1 and 3

anonymous · Answer

I don't even see how L'hopital is applicable here?

anonymous · Answer

it isn't

anonymous · Answer

You write:
As a we have learned/appears-on-page{insert num. p.} a limit of a a ratio of bounded function by another f. that tends to +infty is 0. Now as we know |sin x| =< 1 so it applies.

anonymous · Answer

the largest the numerator can be is 3 the denominator goes to infinity as fast as, well as fast as $x^2$

baldymcgee6 · Answer

@satellite73, but we can't have a negative ln either

anonymous · Answer

you are taking the limit as $x\to \infty$ so presumably it is positive

anonymous · Answer

By the way is it a custom to GIVE or NO-TO-GIVE medals for a complete and correct solution ?

baldymcgee6 · Answer

Hold your horses mikael, ;)

anonymous · Answer

No problem amigo !

baldymcgee6 · Answer

@Mikael, so how do I present my work?

anonymous · Answer

You write:
"As a we have learned/appears-on-page{insert num. p.} a limit of a a ratio of bounded function by another f. that tends to +infty is 0. Now as we know |sin x| =< 1 so it applies."

anonymous · Answer

As a teacher I always would give full mark for solution that is the simplest and shortest of all the CORRECT solutions. This answer satisfies these requirements

baldymcgee6 · Answer

You're a teacher?

anonymous · Answer

and more

baldymcgee6 · Answer

aren't we all. . .

anonymous · Answer

well - what is it that you say ? Do you think there is some common sense in your sequence of remarks and requests ? BTW don't bother - this is a retorical question

baldymcgee6 · Answer

I was simply implying that we are all teachers in our own sense. @Mikael