Find \lim_{n\to\infty} \sqrt[n]{2n+1}

Question

Can someone please explain how this infinite sequence gets into an $e$ form?

$$a_n = \sqrt[n]{2n+1}$$ $$\lim_{n\to\infty} \sqrt[n]{2n+1} = \lim_{n\to\infty} e^{\ln(2n+1)^{1/ n}} =\lim_{n\to\infty} e^\frac{\ln(2n+1)}{n}.$$

How does the nth root of something become $e$ to the natural log?

Answer 1

Remember the definition and basic properties of logarithms:

$$\text{def.}:\;\;\log_ax=y\iff a^y=x\;,\;\;\log a^n=n\log a$$

and since $\;e^x\,,\,\,\log x\;$ are functions inverse to each other, we get

$$e^{n\log a}=e^{\log a^n}=a^n\implies\sqrt[n] a=a^{1/n}=e^{\frac1n\log a}=e^{\log a^{1/n}}$$

BTW, your limit is way easier, imo, with the squeeze theorem:

$$1\le\sqrt[n]{2n+1}\le\sqrt[n]{2n+2n}=\sqrt[n]4\sqrt[n]n\xrightarrow[n\to\infty]{}1\cdot1=1$$

Answer 2

This is because $e^{\ln x}=x$. If you take the logarithm of both sides, then $$ \begin{align} \ln(e^{\ln x})&=\ln x\\ \ln x&=\ln x. \end{align} $$ In your question, $x=\sqrt[n]{2n+1}$.

Answer 3

$$x^n = \exp(\ln(x^n)),$$

because $y = \exp(\ln(y))$. Indeed, the exponential function is the inverse the Napierian logarithm function ln.

$$ \forall x \geq 0, \ y = \exp(x) \Longleftrightarrow \ln(y) = x.$$