# Thread: hint for limit question

1. ## hint for limit question

I'm puzzled with the following limit: $\lim_{n \rightarrow \infty } {\sqrt[n]{{b}^{ \frac{1}{ 2^{n} } } - 1}}$ and $b > 1$
the exponent is $\frac{1}{2^n}$
It is easy to see that the limit lies in [0,1] interval.
Using online graph plotter I can see that the answer is $\frac{1}{2}$ but no matter what I try I can't reach this answer analytically.
I tried quite a few manipulations like: extrating $\frac{1}{2^{n}}$ factor, geometric sum, representing the formula recursively through $a_n$ and $a_{n+1}$ and some other symbol manipulations.

Can someone point me to the promising direction ?
Thank you

2. ## Re: hint for limit question

Originally Posted by ktotam
I'm puzzled with the following limit: $\lim_{n \rightarrow \infty } {\sqrt[n]{{b}^{ \frac{1}{ 2^{n} } } - 1}}$ and $b > 1$
the exponent is $\frac{1}{2^n}$
It is easy to see that the limit lies in [0,1] interval.
Using online graph plotter I can see that the answer is $\frac{1}{2}$ but no matter what I try I can't reach this answer analytically.
I tried quite a few manipulations like: extrating $\frac{1}{2^{n}}$ factor, geometric sum, representing the formula recursively through $a_n$ and $a_{n+1}$ and some other symbol manipulations.

Can someone point me to the promising direction ?
Thank you
\displaystyle \begin{align*} \lim_{n \to \infty} \sqrt[n]{b^{\frac{1}{2n}} - 1} &= \lim_{n \to \infty} \left[ \left( b^{\frac{1}{2n}} - 1 \right) ^{\frac{1}{n}} \right] \\ &= \lim_{n \to \infty} e^{\ln{ \left[ \left( b^{\frac{1}{2n}} - 1 \right) ^{\frac{1}{n}} \right] }} \\ &= \lim_{n \to \infty} e^{ \frac{\ln{ \left( b^{\frac{1}{2n}} - 1 \right) }}{n} } \\ &= e^{\lim_{n \to \infty} \frac{\ln{ \left( b^{\frac{1}{2n}} - 1 \right) }}{n}} \end{align*}

and this is now in a form where you can apply L'Hospital's Rule.

3. ## Re: hint for limit question

Hi, Prove It

Notice that the exponent of b is $\frac{1}{2^n}$ rather than $\frac{1}{2 \cdot n}$.

Anyway I tried what you suggested with $\frac{1}{2^n}$ but it leads to: $\frac{ln{2} \cdot ln{b} \cdot b^{\frac{1}{2^n}}}{2^n \cdot (b^{\frac{1}{2^n}}-1)}$.
It is more complex expression for n so probably not a way to solve it.
By the way n is discrete variable.

4. ## Re: hint for limit question

Originally Posted by ktotam
Hi, Prove It

Notice that the exponent of b is $\frac{1}{2^n}$ rather than $\frac{1}{2 \cdot n}$.

Anyway I tried what you suggested with $\frac{1}{2^n}$ but it leads to: $\frac{ln{2} \cdot ln{b} \cdot b^{\frac{1}{2^n}}}{2^n \cdot (b^{\frac{1}{2^n}}-1)}$.
It is more complex expression for n so probably not a way to solve it.
By the way n is discrete variable.
Prove It is correct. Also, you missed a negative sign. You have:

\begin{align*}\lim_{n\to \infty} \sqrt[n]{b^{2^{-n}}-1} & = \exp\left(\lim_{n\to \infty} \dfrac{\ln\left(b^{2^{-n}}-1\right)}{n}\right) \\ & = \exp\left(\lim_{n\to \infty}-\dfrac{\ln 2\cdot \ln b \cdot b^{2^{-n}}}{2^n\cdot (b^{2^{-n}}-1)}\right) \\ & = \exp\left(\lim_{n\to \infty}-\dfrac{\ln 2\cdot \ln b \cdot 2^{-n}}{1-b^{-2^{-n}}}\right)\end{align*}

You now have another form where you can apply L'Hospital's Rule again. It doesn't matter if n is discrete or not. L'Hospital's still applies.

5. ## Re: hint for limit question

I verified that after applying L'Hospital's rule second time I get the desired limit.
Thanks you