Question

Find where the sequence converges

Answer 1

We can also apply l'Hôpital's rule by first rewriting the limit as

`\displaystyle \lim_(k\to\infty) \left(1 + \frac4k\right)^k = \lim_(k\to\infty) \exp\left(\ln \left(1 + \frac4k\right)^k\right) = \exp\left(\lim_(k\to\infty) (\ln\left(1+\frac4k\right))/(\frac1k)\right)`

Applying the rule gives

`\displaystyle \lim_(k\to\infty) (\ln\left(1+\frac4k\right))/(\frac1k) = \lim_(k\to\infty) \frac{\left(-\frac4{k^2}\right)/\left(1+\frac4k\right)}{-\frac1{k^2}} = 4 \lim_(k\to\infty) \frac1{1 + 4k} = 4`

so that the overall limit is

`\displaystyle \lim_(k\to\infty) \left(1 + \frac4k\right)^k = \lim_(k\to\infty) \exp(4) = \boxed{e^4}`

Answer 2

Answer:
`\\ \lim\limits_(k \to \infty) (1+(4)/(k))^k =e^4.`

Explanation:

`\displaystyle\\ \lim_(k \to \infty) (1+(4)/(k))^k \\x=(x)/(4) *4\\So,\ \lim_(k \to \infty) (1+(4)/(k))^(k)/(4)*4 \\ \lim_(k \to \infty) ((1+(4)/(k))^(k)/(4) )^4.\\Use\ the\ second\ wonderful\ limit:\\\boxed { \lim_(x \to \infty) (1+(1)/(x))^x=e },\\\\So,\\ \lim_(k \to \infty) (1+(4)/(k))^k =e^4.`