lim(x->0) xsin(1/x) use the squeezing theorem
Always, \(-1 \le\sin \le 1\). Does that give you a proper hint? :)
i don't quite understand...
Have you ever heard of the Squeeze Theorem?(It's also known as 'Sandwich' or 'Pinching' Theorem).
yeah, that's the hint given in the question.
So you haven't heard of it. Hmm. I recommend you to see this link. http://tutorial.math.lamar.edu/Classes/CalcI/ComputingLimits.aspx
wait, i think i know what you mean.
Great!
Can you make an inequality to start with?
im still wondering how @ParthKohli knows the squeeze
Also to remember: \(-1 \le \sin x \le 1 \Longrightarrow {-1 \le \sin{1 \over x} \le 1}\)
@lgbasallote I started Calculus a week ago.
Heh - no need to worry! This is not a copied answer!
but this is not "just calculus"
This is a part of limits, and I am studying Calculus on Paul's Notes.
the answer's 0 right?
what about lim(x->0+) (lnx)/(cotx)
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