Q118 / 200 · Manipal (MET) 2013 · Mathematics

Figure

  • A
    $\log (a b)$
  • B
    $\log (a / b)$
  • Correct answer
    C
    $\log (b / a)$
  • D
    $-\log (a / b)$

Solution

The given limit $\begin{aligned} & L=\lim _{n \rightarrow \infty}\left\{\frac{1}{n a}+\frac{1}{n a+1}+\frac{1}{n a+2}+\ldots+\right. \\ &\left.\qquad \frac{1}{n a+n(b-a)}\right\}\end{aligned}$ $\begin{aligned} & =\lim _{n \rightarrow \infty} \sum_{r=0}^{n(b-a)} \frac{1}{n a+r} \\ & =\lim _{n \rightarrow \infty} \frac{1^{(b-a) n}}{\sum_{r=0}^{(b-a} \frac{1}{a+r / n}} \\ & =\int_0^{(b-a)} \frac{d x}{a+x}=[\log (a+x)]_0^{b-a} \quad\left(\frac{r}{n}=x\right) \\ & =\log b-\log a \\ & =\log \left(\frac{b}{a}\right)\end{aligned}$

Hence, the correct answer is option: 'C'.

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