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Another tricky case where we cannot determine a limit is when a function oscillates infinitely fast as it approaches a point. We call such oscillations indeterminate oscillations. Consider the function

f(x)=\sin (x^{-1})

as x approaches zero from the right. The reciprocal of x is not continuous at zero because division by zero is not defined. To the right of zero, f(x) is continuous but

\sin (1/0^{+})

can’t be evaluated.


Example 1. Determine the limit

\mathop{\lim }\limits_{x\to 0^{+} } \sin (x^{-1} )

Solution 1. As x approaches zero from the right then the inverse of x approaches infinity. This limit is the same as writing

\mathop{\lim }\limits_{x\to 0^{+} } \sin (x^{-1} )=\mathop{\lim }\limits_{u\to \infty } \sin u

I have made the substitution u=1/x. We know that the sine of infinity does not exist because the function is always oscillating there. Therefore the original limit does not exist either.

\mathop{\lim }\limits_{x\to 0^{+} } \sin (x^{-1} )=DNE

../images/S1OX.pdf

We see that the function $\sin (x^{-1})$ does not converge as $x$ approaches zero from the right

Similarly a random continuous function has no limit at infinity because it doesn’t converge to any value. This is not because of indeterminate oscillations, but rather a random walk.


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