What is the limit of sinx/x as x approaches infinity?

[helloe]

I don't know how to solve this... please help! Thanks in advance!!

 

[funny]

we know that -1 less than or equal to sinx less than or equal to 1
==> for large positive x, [-1/x]less than or equal to sinx/x less than or equal to 1/x
now, by applying x tends to infinity,
0 less than or equal to limx-->infinity [sinx/x]less than or equal to 0
hence, the answer is zero

 

[AlephNull]

The limit is 0.
As x tends to infinity, you are dividing something that can only be between 1 and -1 by an increasingly large number... which tends to 0

 

[CogitoErgoCogitoSum]

0
zero

You can solve this with the Squeeze Theorem. But personally, I think the proof is rather intuitive.

Sine function, for all real inputs 'x' in the domain, is guaranteed to be stuck between -1 and +1. Right? -1 sin(x) 1.

That being the case, as x approaches infinity, the numerator will always be somewhere in that range. Its specific value is irrelevant.

Notice the denominator. It constantly grows without bound. The numerator is essentially a constant with respect to the infinitely large denominator

 

[PlayerX]

Should be 0