-1
$\begingroup$

I have to find limit of the following sequence:

Suppose $a>0$, consider the sequence $a_n=n\left(\sqrt[n]{ea}- \sqrt[n]{a}\right),\,n\ge1 \,\text{then} \lim_{n \to \infty}a_n\,\text{is}\ldots$

I tried to solve this problem and considered $a$ to be a very small positive number and I got the result that sequence converges to $0$.

But actually the answer is that sequence converges to 1.

How shall I proceed to find limit of this sequence.

  • 0
    Have you already covered functions, their limits, derivatives ,etc.? – 2017-01-06
  • 0
    "I got the result that sequence converges to 0" You could explain precisely what you did to get this. – 2017-01-06

3 Answers 3

3

One may write, as $n \to \infty$, $$ \begin{align} a_n&=n\left(e^{\frac{\large\ln(ea)}n}-e^{\large\frac{\ln(a)}n} \right) \\&=n\left(e^{\large\frac{1+\ln(a)}n}-e^{\large\frac{\ln(a)}n} \right) \\&=n \cdot e^{\large\frac{\ln(a)}n}\left(e^{\large\frac{1}n}-1 \right) \\&= e^{\large\frac{\ln(a)}n}\cdot\frac{e^{\frac{1}n}-1}{\frac{1}n} \end{align} $$ then one may conclude with $$ \lim_{n \to \infty}\frac{e^{\frac{1}n}-1}{\frac{1}n}=1. $$

  • 0
    @AbhishekChandra You are welcome. I have a lot of practice ;) – 2017-01-06
0

$$n\left(\sqrt[n]{ea}-\sqrt[n]a\right)=n\sqrt[n]a\left(\sqrt[n]e-1\right)=\sqrt[n]a\,\frac{\sqrt[n]e-1}{\frac1n}\xrightarrow[n\to\infty]{}1\cdot\left[\left(e^x\right)'\right]_{x=0}=1\cdot e^0=1$$

0

The function $$A:[1,+\infty)\to\Bbb R: x\mapsto xa^{\frac{1}{x}}(e^{\frac{1}{x}}-1)$$ is continuous as a product of composite functions of continuous functions.

Hence, if the limit when $x\to+\infty$ exists, it is the same along any sequence $(x_{n})_{n}$ such that $x_{n}\to+\infty$ as $n\to+\infty$. We shall prove that the limit exists and thus for the sequence $x_{n}=n$ as well.

We have:

\begin{align*} &\lim_{x\to+\infty}A(x)\\ &=\lim_{x\to+\infty}xa^{\frac{1}{x}}(e^{\frac{1}{x}}-1)\\ &=\lim_{x\to+\infty}\frac{a^{\frac{1}{x}}(e^{\frac{1}{x}}-1)}{1/x} \end{align*}

Here, we have a $\frac{0}{0}$ limit and we can thus use L'Hospital's rule. It yields:

\begin{align*} &\lim_{x\to+\infty}A(x)\\ &=\lim_{x\to+\infty}\frac{a^{\frac{1}{x}}(e^{\frac{1}{x}}-1)}{1/x}\\ &=\lim_{x\to+\infty}\frac{\ln(a)a^{\frac{1}{x}}\left(-\frac{1}{x^{2}}\right)(e^{\frac{1}{x}}-1)+a^{\frac{1}{x}}\left(-\frac{1}{x^{2}}\right)e^{\frac{1}{x}}}{-\frac{1}{x^{2}}}\\ &=\lim_{x\to+\infty}\ln(a)a^{\frac{1}{x}}(e^{\frac{1}{x}}-1)+e^{\frac{1}{x}}a^{\frac{1}{x}}\\ &= 1 \end{align*}