"Prove that: limn→∞(1+1an)an=e if limn→∞an=∞" was answered by students like you

iocasq4

Answered question

2022-01-23

Prove that:

lim_{n \to \infty} {(1 + \frac{1}{a_{n}})}^{a_{n}} = e

lim_{n \to \infty} a_{n} = \infty

Ydaxq

Beginner2022-01-24Added 12 answers

Given that

lim_{n \to \infty} b_{n} = lim_{n \to \infty} \frac{1}{a_{n}} = 0

we have,

lim_{n \to \infty} {(1 + \frac{1}{a_{n}})}^{a_{n}} = lim_{n \to \infty} \exp (\frac{\ln (1 + \frac{1}{a_{n}})}{\frac{1}{a_{n}}}) = lim_{h \to 0} \exp (\frac{\ln (1 + h)}{h}) = e

similarly

lim_{n \to \infty} {(1 + b_{n})}^{\frac{1}{b_{n}}} = lim_{n \to \infty} \exp (\frac{\ln (1 + b_{n})}{b_{n}}) = lim_{h \to 0} \exp (\frac{\ln (1 + h)}{h}) = e

fumanchuecf

Beginner2022-01-25Added 21 answers

For

| t | < 1

, note that

t - \frac{1}{2} t^{2} \leq \log (1 + t) \leq t

and so

| \log (1 + t) - t | \leq {\frac{12}{t}}^{2}

Hence for

| x | > 1

we have

| \log (1 + \frac{1}{x}) - \frac{1}{x} | \leq {\frac{12}{\frac{1}{x}}}^{2}

and so

| x \log (1 + \frac{1}{x}) - 1 | \leq \frac{12}{\frac{1}{| x |}}

Hence

lim_{x \to \infty} x \log (1 + \frac{1}{x}) = lim_{x \to \infty} {\log (1 + \frac{1}{x})}^{x} = 1

from which it follows that

lim_{x \to \infty} {(1 + \frac{1}{x})}^{x} = e

RizerMix

Expert2022-01-27Added 656 answers

Lets

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