5
$\begingroup$

This question already has an answer here:

$$\lim_{n\to \infty}2^{\frac{1}{n}}$$

Basically looking at $$\lim_{n\to \infty}2^{\frac{1}{n}}=2^{\lim_{n\to \infty}{\frac{1}{n}}}=2^0=1$$

But if I use $$\lim_{n\to \infty}2^{\frac{1}{n}}=\lim_{n\to \infty}e^{\ln\left(2^{\frac{1}{n}}\right)}=\lim_{n\to \infty}e^{{\frac{1}{n}}\cdot{\ln(2)}}=\lim_{n\to \infty}e^{\frac{1}{n}}\cdot\lim_{n\to \infty}e^{\ln(2)}=1\cdot2=2$$

Where did I got it wrong?

$\endgroup$

marked as duplicate by user147263, HK Lee, Tomás, John B, Shailesh Apr 11 '16 at 0:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ The calculation is incorrect, $\lim e^{\frac{1}{n} \log 2} =e^0=1$ since $(\log 2)/n \to 0$ as $n\to \infty$, where $\log x$ is the natural logarithm. $\endgroup$ – Nap D. Lover Mar 20 '16 at 8:05
  • 6
    $\begingroup$ When you said that $e^{\frac1n\log2}=e^{\frac1n}e^{\log2}$. Actually, $e^ae^b=e^{a+b}$, not $e^{ab}$. $\endgroup$ – Did Mar 20 '16 at 8:05
5
$\begingroup$

Rules of exponentiation:

  • $a^{b+c}=(a^b)\times(a^c)$
  • $a^{b\times c}=(a^b)^c$

Therefore:

  • $\lim\limits_{n\to\infty}e^{{\frac{1}{n}}+{\ln2}}=\left(\lim\limits_{n\to\infty}e^{\frac{1}{n}}\right)\times\left(\lim\limits_{n\to\infty}e^{\ln2}\right)$
  • $\lim\limits_{n\to\infty}e^{{\frac{1}{n}}\times{\ln2}}=\left(\lim\limits_{n\to\infty}e^{\frac{1}{n}}\right)^{\lim\limits_{n\to\infty}\ln2}$
$\endgroup$
4
$\begingroup$

$$\lim_{n\to \infty} e^{\frac{1}{n}\cdot \ln 2} \neq \displaystyle \lim_{n \to \infty} e^{\frac{1}{n}}\cdot \displaystyle \lim_{n \to \infty} e^{\ln2}$$

$\endgroup$
3
$\begingroup$

Personally I prefer to do it this way

$$\lim_{n\to \infty} e^{\frac{1}{n}\cdot\ln2}=\lim_{n\to \infty}\bigl( e^{\ln2}\bigr)^{\frac{1}{n}}=\bigl(e^{\ln2}\bigr)^{\lim_{n\to \infty}\frac{1}{n}}=\bigl(e^{\ln2}\bigr)^0=1$$

But maybe thats just me. the other ways work too

$\endgroup$
2
$\begingroup$

See $\lim \limits_{n - \infty}e^{\frac{1}{n}*\ln2}=\lim \limits_{n - \infty}(e^{\frac{1}{n}})^{\ln2}=1^{\ln2}=1$

$\endgroup$
1
$\begingroup$

Let $\alpha_n$ be a convergent sequence with non-negative terms such that $$2^{1/n}=1+\alpha_n.$$

Then, we have, $$2=(1+\alpha_n)^n=1+n\alpha_n+\frac {n(n-1)}2\alpha_n^2+\dots\\\implies2\gt1+n\alpha_n\\\implies\alpha_n\lt \frac 1n$$

So, we have $$0\lt \alpha_n\lt \frac 1n.$$

So, $$\alpha_n\to 0.$$

So, $$\lim 2^{1/n}=\lim(1+\alpha_n)=1+\lim \alpha_n=1+0\\\implies \lim2^{1/n}=1$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.