Exploring $\lim_{n\rightarrow \infty }(2-2^{\frac{1}{n+1}})^{n}$

• MHB
• Vali
In summary, the limit given is equal to e raised to the power of the limit as n approaches infinity of n times the quantity 1 minus 2 to the power of 1 over n plus 1. This can be solved using the remarkable limit $\lim_{x\rightarrow 0}(1+x)^{\frac{1}{x}}=e$. After applying L'Hopital's rule, the limit is found to be equal to $\frac{1}{2}$.
Vali
$$\lim_{n\rightarrow \infty }(2-2^{\frac{1}{n+1}})^{n}$$
I wrote that $L=e^{\lim_{n\rightarrow \infty }(n)(1-2^{\frac{1}{n+1}})}$
how to continue ?

What I would do is write:

$$\displaystyle L=\lim_{n\to\infty}\left(\left(2-2^{\frac{1}{n+1}}\right)^n\right)$$

And this implies:

$$\displaystyle \ln(L)=\lim_{n\to\infty}\left(\frac{\ln\left(2-2^{\frac{1}{n+1}}\right)}{\dfrac{1}{n}}\right)$$

Now you have the indeterminate form 0/0.

After I apply L'Hopital I get $ln(L)=-ln(2)$ so $L=\frac{1}{2}$
One question,Can this limit be solved by using this remarkable limit? $\lim_{x\rightarrow 0}(1+x)^{\frac{1}{x}}=e$
Thank you very much!

1. What is the limit of the given expression as n approaches infinity?

The limit of the expression is 1.

2. How do you solve this limit problem?

To solve this limit problem, we can use the limit laws and algebraic manipulations to rewrite the expression in a simpler form. Then, we can use the substitution method to evaluate the limit.

3. What is the significance of the exponent in the expression?

The exponent in the expression, n+1, represents the number of times the base, 2, is multiplied by itself. As n approaches infinity, the exponent becomes larger and the value of the expression approaches 1.

4. Can this limit be evaluated using L'Hopital's rule?

No, L'Hopital's rule can only be used for indeterminate forms such as 0/0 or ∞/∞. This expression does not fall under those categories and therefore cannot be evaluated using L'Hopital's rule.

5. How does the value of the limit change as n increases?

As n increases, the value of the limit approaches 1. This can be seen by plugging in larger values for n and observing that the expression gets closer and closer to 1.

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