Question
Let $C_{n}=\int_{1 / 2+1}^{1 / 8} \frac{\tan ^{-1}(n x)}{\sin ^{-1}(n x)} d x$, then $\lim _{n \rightarrow-} n^{2} \cdot C_{n}$ is equalto(a) 1(b) 0(c) $-1$(d) $\frac{1}{2}$
Step 1
We can rewrite this as $C_{n}=\frac{1}{n}\int_{n(1/2+1)}^{n(1/8)} \frac{\tan ^{-1}(t)}{\sin ^{-1}(t)} dt$ by substituting $t=nx$. Show more…
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