How to find this limit ? I've tried this problem but in the middle steps i'm stuck. I'll upload the image of the solution
Asked
Active
Viewed 75 times
0
Michael Rozenberg
- 194,933
Priya Wadhwa
- 173
-
I think the $1/x$ is in exponent and not as a factor. If it is a factor then the problem is trivial as first factor tends to $1$ and second factor tends to $\pm\infty$. If $1/x$ is in exponent then just see this answer https://math.stackexchange.com/a/1849862/72031 – Paramanand Singh Jan 06 '18 at 03:20
2 Answers
0
It's $$\lim_{x\rightarrow0}\left(1+\frac{1^x+2^x+...+n^x}{n}-1\right)^{\frac{n}{1^x-1+2^x-2+...+n^x-n}\cdot\frac{\frac{2^x-1}{x}+...+\frac{n^x-1}{x}}{n}}=e^{\frac{\ln2+...+\ln n}{n}}=\sqrt[n]{n!}.$$ I used the following.
1.For all $a>0$ we have $$\lim_{x\rightarrow0}\frac{a^x-1}{x}=\ln{a}.$$
- Since $\frac{1^x+2^x+...+n^x}{n}-1\rightarrow0$ for $x\rightarrow0$ and $\lim\limits_{x\rightarrow0}(1+x)^{\frac{1}{x}}=e$, we obtain $$\lim_{x\rightarrow0}\left(1+\frac{1^x+2^x+...+n^x}{n}-1\right)^{\frac{n}{1^x-1+2^x-2+...+n^x-n}}=e$$
- If there is $\lim\limits_{x\rightarrow a}f(x)=A,$ where $A>0$ and there is $\lim\limits_{x\rightarrow a}g(x)=B$ then there is $$\lim_{x\rightarrow a}f(x)^{g(x)}=A^B.$$
Michael Rozenberg
- 194,933
-
-
-
-
-
Duplicate of https://math.stackexchange.com/questions/1849697/whats-the-proof-for-lim-x-%E2%86%92-0-a-1x-a-2x-a-nx-n1-x-a?noredirect=1&lq=1 – Priya Wadhwa Jan 06 '18 at 05:56