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fine the limit :

$$\lim_{n\to \infty} (1+\frac{\sqrt[n]{a}-1}{b})^n=?$$

$a,b\in \mathbb{R}$

My Try :

$$\lim_{n\to \infty} (1)+\lim_{n\to \infty}(\frac{\sqrt[n]{a}-1}{b})^n=1+0=1$$

Because:

$$\lim_{n\to \infty} \sqrt[n]{a}=1 $$

is it right ?

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    You need to see that $(x+y)^n$ not always is equal to $x^n+y^n$. You hace an indetermination of the form $1^ \infty$. Find the limit of the natural logarithm, and finally you need to apply exponential function. – 2017-02-25
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    @asd . $(\lim_{n\to \infty}(\frac{\sqrt[n]{a}-1}{b}))^n=\lim_{n\to \infty}(\frac{\sqrt[n]{a}-1}{b})^n$ – 2017-02-25
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    @Almot1960 You start by writing $\lim_n (1+f(n))^n = \lim_n 1+\lim_n (f(n)^n)$.As asd points out, this is not correct. – 2017-02-25
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    @Almot1960 Your tag is "without L'Hospital". Is this intended to be a problem without the Rule? – 2017-02-25

2 Answers 2

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No, this is not right.

  • Your first step: $$ \lim_{n\to \infty} (1+\frac{\sqrt[n]{a}-1}{b})^n = \lim_{n\to \infty} 1 + \lim_{n\to \infty} (\frac{\sqrt[n]{a}-1}{b})^n $$ is wrong. With the same argument, you would have $$\lim_{n\to\infty} (1+1)^n = \lim_{n\to\infty} 1^n + \lim_{n\to\infty} 1^n = 1+1 = 2$$ which is clearly false.

  • Now, let us prove that the limit is $a^{1/b}$. For everything to make sense to begin with, one must have $a,b>0$. I will only use elementary arguments involving the fact that $\frac{d}{dx}e^x\mid_{x=0} = 1$ and $\frac{d}{dx}\ln(1+x)\mid_{x=0} = 1$.

    To begin with, let us put our quantity is the simpler to analyze exponential form: $$ (1+\frac{\sqrt[n]{a}-1}{b})^n = e^{n \ln\left(1+\frac{\sqrt[n]{a}-1}{b}\right)} = \exp\left({n \ln\left(1+\frac{e^{\frac{\ln a}{n}}-1}{b}\right)}\right) $$ so by continuity of $\exp$, it suffices to find the the limit of ${n \ln\left(1+\frac{e^{\frac{\ln a}{n}}-1}{b}\right)}$.

    Since $\frac{e^{\frac{\ln a}{n}}-1}{b}\xrightarrow[n\to\infty]{0}$ and $\lim_{x\to 0} \frac{\ln(1+x)}{x} = 1$, we can rewrite $$ {n \ln\left(1+\frac{e^{\frac{\ln a}{n}}-1}{b}\right)} = n \cdot \frac{e^{\frac{\ln a}{n}}-1}{b} \cdot \underbrace{ \frac{\ln\left(1+\frac{e^{\frac{\ln a}{n}}-1}{b}\right)}{\frac{e^{\frac{\ln a}{n}}-1}{b}}}_{\to_{n\to\infty} 1} $$ and thus focus on the limit of $n \cdot \frac{e^{\frac{\ln a}{n}}-1}{b}$. Recalling that $1=\exp'(0) = \lim_{x\to 0} \frac{e^x-1}{x}$ and that $\frac{\ln a}{n}\xrightarrow[n\to\infty]{} 0$, we then can (!) write $$n \cdot \frac{e^{\frac{\ln a}{n}}-1}{b} =n\cdot\frac{\frac{\ln a}{n}}{b} \cdot \frac{e^{\frac{\ln a}{n}}-1}{\frac{\ln a}{n}} = \frac{\ln a}{b} \cdot \frac{e^{\frac{\ln a}{n}}-1}{\frac{\ln a}{n}} \xrightarrow[n\to\infty]{}\frac{\ln a}{b}\cdot 1. $$ Putting it all together, $$ (1+\frac{\sqrt[n]{a}-1}{b})^n = \exp\left({n \ln\left(1+\frac{e^{\frac{\ln a}{n}}-1}{b}\right)}\right) \xrightarrow[n\to\infty]{} \exp\left(\frac{\ln a}{b}\cdot 1\cdot 1\right) = e^{\frac{\ln a}{b}} $$ giving that the limit is $e^{\frac{\ln a}{b}}=\boxed{a^{\frac{1}{b}}}$.

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    If you are familiar and comfortable with Taylor approximations, the whole second part can become a much more concise 3-line derivation. – 2017-02-25
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    I know your answer is right, but I made an error somewhere in my answer and I can't find it. Could you look it over? – 2017-02-25
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    @SimplyBeautifulArt You do get the same result, though -- don't you? – 2017-02-25
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    Check my last few steps and you'll see I sneaked the result through. But I want to fix it :-/ – 2017-02-26
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    I'll have a look asap! (Currently outside under the rain.) – 2017-02-26
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    :O May your future be not rained upon then – 2017-02-26
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    @SimplyBeautifulArt Sorry to be daft, but your derivation looks fine to me. Which step is the one bugging you? (Showing the last limit is 1?) – 2017-02-26
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    Look really closely at the last limit. – 2017-02-26
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    Since $\sqrt[n]{a}\to 1$, this goes to $\frac{1}{1+\frac{1-1}{b}} = 1$ -- or am I misunderstanding what you say? – 2017-02-26
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    OH! Wow, I'm stupid. I was thinking that $\sqrt[n]a\to\infty$! XD Thank you for your helps – 2017-02-26
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    No worries! ${}{} $ – 2017-02-26
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    You need $a>0, b\ne 0$ in this problem. – 2017-02-27
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    Yes, indeed... i wrote a,b > 0 for comvenience. – 2017-02-27
2

Let $L$ be the limit. Then,

$$\ln(L)=\lim_{n\to\infty}\frac{\ln\left(1+\frac{\sqrt[n]a-1}b\right)}{1/n}\stackrel{L'H}=\lim_{n\to\infty}\frac{(-1/n^2)\frac{\ln(a)}b\sqrt[n]a}{\left(1+\frac{\sqrt[n]a-1}b\right)(-1/n^2)}\\=\frac{\ln(a)}b\lim_{n\to\infty}\frac1{\left(\frac1{\sqrt[n]a}+\frac{1-\frac1{\sqrt[n]a}}b\right)}\\=\frac{\ln(a)}b$$

Thus,

$$L=e^{\ln(a)/b}=\sqrt[b]a$$

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    I am not sure L'Hopital's rule is completely fine by the OP, given the tag. – 2017-02-25
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    @ClementC. Of course, but still, it's probably the recommended path for this type of problem. – 2017-02-26
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    You are assuming the limit exists at the outset. The above will be better written without all the limit notation until the end. – 2017-02-27