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DIA [1.3K]
1 year ago
5

Find where the sequence converges

Mathematics
2 answers:
soldier1979 [14.2K]1 year ago
8 0

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}

avanturin [10]1 year ago
3 0

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.

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