We know that \begin{equation} \lim_{t \to 0} \frac{\sin(t)}{t}=1. \end{equation} I was wondering what would be \begin{equation} \lim_{t \to 0} \frac{\sin(\ln t)}{t}. \end{equation} I tried using L'hospital's rule, but I am not getting anywhere with that. Any help will be deeply acknowledged.
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3 Answers
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You may choose $t_{n}>0$ such that $t_{n}\rightarrow 0$ and that $\ln t_{n}=-2n\pi+\pi/6$, then $\sin(\ln t_{n})=1/2$ and hence the limit is $\infty$.
On the other hand, if you choose $s_{n}>0$ such that $s_{n}\rightarrow 0$ and that $\ln s_{n}=-2n\pi$, then $\sin(\ln s_{n})=0$, the limit is $0$.
The limit does not exist in the extended real sense.
answered Apr 18, 2018 at 4:50
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HINT:
Put $t = e^x$.
Also, notice L.H.L doesn't exist
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The limit does not exist as @saulspatz said because ln(x) is not defined for x<0 and x=0. That's what we call Left Hand Limit. However, you might be interested in computing the RHL if it exists. @user284331 explained that by choosing distinct specific sequences of real numbers, say t_n, you may obtain different values when calculating the limit. This contradicts the uniqueness of the limit and the RHL thus does not exist.
