Question

How can I solve: sin(ln(x+1000)-sin(ln(x)) when x approaches infinity

Answer 1

Use the Mean Value Theorem for
\[
g(x)=\sin (\ln x) \text { in the interval } [x, 1000+x]
\]

Answer 2

There is a \(c_x\ \in [x,1000+x]\) such that \[
g(x+1000)-g(x) = 1000 g'(c_x) \\
|\sin (\ln( x + 1000) -\sin(\ln x)| =| 1000 \
\frac{\cos(\ln( c_x)} {c_x}|\le \\ 1000 \frac 1 {x_c}\le 1000 \frac 1 {x}
\]
Can you finish it now?

Answer 3

Sure!! Your response is awesome. Now I know how to deal with the problem.

Answer 4

YW