### CSIR JUNE 2011 PART C QUESTION 65 SOLUTION ($\text{lim}_{n \to \infty} n\,\text{log}(\frac{1}{n+1})$))

Which of the following is/are correct?
1) $n\,log(1+\frac{1}{n+1}) \to 1$ as $n \to \infty$,
2) $(n+1)\,log(1+\frac{1}{n}) \to 1$ as $n \to \infty$,
3) $n^2\,log(1+\frac{1}{n}) \to 1$ as $n \to \infty$,
4) $n\,log(1+\frac{1}{n^2}) \to 1$ as $n \to \infty$.
Solution

option a:(True) $n\,log(1+\frac{1}{n+1}) = log((1+\frac{1}{n+1})^n) =log(\frac{(1+\frac{1}{n+1})^{n+1}}{(1+\frac{1}{n+1})})$.
$lim_{n \to \infty} log(\frac{(1+\frac{1}{n+1})^{n+1}}{(1+\frac{1}{n+1})}) = log(\lim_{n \to \infty} \frac{(1+\frac{1}{n+1})^{n+1}}{(\frac{1}{n+1})} )$ [Since log is a continuous function]. Now, the numerator converges to $e$ as $\lim_{n \to \infty}(1+\frac{1}{n})^n = e$ and the denominator converges to $1$, so the ratio converge to $e$. So the limit is $log(e) = 1$.
option b:(True) $(n+1)(log(1+\frac{1}{n})) = log(1+\frac{1}{n})^{n+1} = log((1+\frac{1}{n})^n (1+\frac{1}{n}))$. Taking limit $n \to \infty$, we see that the inside limit is $e \cdot 1 = e$ and the required limit is $log(e) = 1$.
option c: (False) $n^2\,log(1+\frac{1}{n}) = \frac{log(1+\frac{1}{n})}{\frac{1}{n^2}} = \frac{0}{0} \text{form}$. We can use L'Hospital's rule. Differentiate the numerator and the deminator we get $\frac{(\frac{n}{n+1})(\frac{-1}{n^2})}{\frac{-2}{n^3}} = \frac{n^2}{2(n+1)}$ which is divergent as $n \to \infty$ and hence the required limit doesn't exists
option d: (False) $n log(1+\frac{1}{n^2}) = \frac{log(1+\frac{1}{n^2})}{\frac{1}{n}} = \frac{0}{0} \text{ form}$. Again, we can use the L'Hospital's rule. Differentiate the numerator and the denominator we get $\frac{(\frac{n^2}{1+n^2})(\frac{-2}{n^3})}{\frac{-1}{n^2}} = (\frac{n^2}{1+n^2})(\frac{2}{n})$ which converges to $0$ as $n \to \infty$. Hence the required limit is $log(0)$ which doesn't exist.
### NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
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