Question

Find where the sequence converges

Answer 1

Answer:  $\\ \lim\limits_{k \to \infty} (1+\frac{4}{k})^k =e^4.$

Step-by-step explanation:

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

Answer 2

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

$\displaystyle \lim_{k\to\infty} \left(1 + \frac4k\right)^k = \lim_{k\to\infty} \exp\left(\ln \left(1 + \frac4k\right)^k\right) = \exp\left(\lim_{k\to\infty} \frac{\ln\left(1+\frac4k\right)}{\frac1k}\right)$

Applying the rule gives

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

so that the overall limit is

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