Question

How can I find a limit of ln(sin(1/x))+ln(x)?

Answer 1

What is the variable x going towards in the limit?

Answer 2

\( \displaystyle \lim_{x \to ???} \ln \left( \sin \left(1/x\right) \right) + \ln x \)

Answer 3

when x approaches to infinity

Answer 4

Hm.. interesting limit.
I think we could start by reducing this into terms of a single logarithm using the multiplication property:
\( \log A + \log B = \log (AB) \)

Answer 5

then what should i do next? i've already reached that step and got stuck

Answer 6

So you have:
\( \displaystyle \lim_{x \to \infty} \ln \left( x \sin (1/x) \right) \)
This looks suspiciously like the sin x / x limit as x approaches 0. Have you seen a rule about substitution like this? \( x = \dfrac{1}{u} \), \( x \to \infty \) as \(u \to 0+ \)

Answer 7

So that we have
\( \displaystyle \lim_{u \to 0+} \ln \left( \dfrac{\sin u}{u} \right) \)
The interior of the limit would now be approaching 1 (sin x / x limit), so the logarithm itself is approaching ln(1) = 0.