Limit of a sine function problem.

[Checkfate]

Part a)I need help understanding a step in solving \lim_{x\rightarrow \infty}xsin(\frac{1}{x})

The textbook is suggesting I replace \frac{1}{x} with y, so that I can get a limit in the form

\lim_{y\rightarrow 0}\frac{siny}{y} which is understandably easy to solve. The part I don't understand is how the limit changes from x approaches inf to y approaches 0. I was running into the same confusion with this problem as well so perhaps I should post it.

Part b)Solve \lim_{x\rightarrow 0}\frac{sin4x}{sin3x}

They then proceed in their next step to write it as \frac{\lim_{4x\rightarrow 0}4x(\frac{sin4x}{4x})}{\lim_{3x\rightarrow 0}3x(\frac{sin3x}{3x})}

How do they get the 4x approaches 0 and 3x approaches 0, I understand it as a "this is the way to do it" and can do it easily, but I'd like to know what they are doing. Anyways, thanks

 

[Office_Shredder]

Forthe first question: If x is approaching infinity, what is 1/x approaching? Therefore, what is y approaching?

For the second question, I can't see your latex image, but I think I can help you with your question. If x approaches zero, what would you think 4x approaches? Specifically, try taking the limit of 4x as x goes to zero, and remember that if there's no problem, as a general rule of thumb you can just plug in what x approaches and do the limit real easy

 

[StatusX]

If f(x)->b as x->a, then the limit of g(f(x)) as x goes to a is equal to the limit of g(y) as y goes to b. Proving this is a nice exercise in the epsilon delta definition of a limit.

 

[Checkfate]

Thankyou both, :D