Limits for Sin(1/x^n)

1. Oct 23, 2008

janakiraman

Hi

I have been trying to work with limits and i came across a problem where it was defined that the limit for Sin(1/x) is not defined at x=0. Well i completely understand that part because the function oscillates for any small limit near 0. But i don't really understand how to calculate the limit for x for example at 0.1. Because the function still oscillates and what would be the ideal limit value that i need to take so that the function will proceed towards value sin(1/0.1)? And how to calculate similarly for sin(1/x^2), sin(1/x^3) etc until Sin(1/x^n) for x=0.1 from the definition of limits.

I tried to do a matlab code and generate graph to see if there is a pattern in oscillation, but unfortunately i could not decode anything. I have attached the graph here, the blue curve indicates sin(1/x), red for sin(1/x^2), green for sin(1/x^3) and black for sin(1/x^4).

P.S: In my opinion i felt this problem does not fall into homework questions, but i'm extremely sorry if someone else feels that way

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2. Oct 23, 2008

HallsofIvy

Sin(x) is a continuous function so
$$\lim_{x\rightarrow x_0} sin(\frac{1}{x})= sin(\lim_{x\rightarrow x_0}\frac{1}{x})$$
That inner limit exist for all x0 except 0. The limit as x approaches 0.1 is just sin(1/0.1)= sin(10).

If x is close enough to 0.1, for example if 1/(0.1+ $\pi$/2)< x< 1/(0.1- $\pi$/2) on your graph (x, sin(1/x) ) is on ONE of the many oscillations shown on your graph and the others don't matter.

3. Oct 23, 2008

janakiraman

hey thank you very much. infact even i realized the mistake in my question after i asked it :). But another question is the way you took the limit inside sin is possible for all continuous curves right?

4. Oct 23, 2008

HallsofIvy

Yes, that's pretty much the definition of continuous!