Saturday, 20 June 2015

calculus - I need to understand why the limit of xcdotsin(1/x) as x tends to infinity is 1



here's the question, how can I solve this:



lim



Now, from textbooks I know it is possible to use the following substitution x=1/t, then, the ecuation is reformed in the following way



\frac{\sin t}{t}




then, and this is what I really can´t understand, textbook suggest find the limit as t\to0^+ (what gives you 1 as result)



Ok, I can't figure out WHY finding that limit as t approaches 0 from the right gives me the answer of the limit in infinity of the original formula. I think I can't understand what implies the substitution.



Better than an answer, I need an explanation.



(Sorry If I wrote something incorrectly, the english is not my original language)
Really thanks!!


Answer




\lim_{x\to\infty}x\sin(1/x) is like the limit of this sequence: 1\sin(1/1), 10\sin(1/10), 100\sin(1/100),\ldots where we have inserted x marching along from 1 to 10 to 100 on its merry way to \infty. This is almost literally the same as \frac{1}{1}\sin(1), \frac{1}{0.1}\sin(0.1), \frac{1}{0.01}\sin(0.01),\ldots which is an interpretation of \lim_{t\to0^+}\frac{1}{t}\sin(t) with t marching its way from 1 down to 0.1 down to 0.01 on its merry way to 0. So whatever the value of these limits are, \lim\limits_{x\to\infty}x\sin(1/x)=\lim\limits_{t\to0^+}\frac{1}{t}\sin(t).


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