-1

https://www.symbolab.com/solver/limit-calculator/%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cleft%28%5Cfrac%7Bn%7D%7Bn%2B1%7D%5E%7Bn%7D%5Cright%29/?origin=enterkey

Dat should be simple but it don't work. And symbolab don't give nothing, just formula that insta solves it but I just know that if I used it I'll get owned, no points, no money.

Do I use lupitall, I tried it gets messy idk.

So how do I really solve it?

E.H.E
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user779444
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2 Answers2

6

$$(\frac{n}{n+1})^n=\frac{1}{(1+\frac{1}{n})^n}$$

E.H.E
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0

Let $$\displaystyle y=\lim_{n\rightarrow \infty}\left(\frac{n}{n+1}\right)^n\;,$$ Now taking $\ln$ on both side, we get

$$\displaystyle \ln(y) = \lim_{n\rightarrow \infty}n\cdot \ln\left(\frac{n}{n+1}\right)$$

Now Put $\displaystyle n=\frac{1}{m}\;,$ So $$\displaystyle \ln(y)=\lim_{m\rightarrow 0}\frac{\ln\left(\frac{1}{m+1}\right)}{m} = -\lim_{m\rightarrow 0}\frac{\ln(m+1)}{m} =-1$$

above we have used the formula $$\displaystyle \bullet \lim_{x\rightarrow 0}\frac{\ln(1+x)}{x} =1$$

So we get $$\displaystyle \ln(y) = -1\Rightarrow y=e^{-1}=\frac{1}{e}$$

Akiva Weinberger
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juantheron
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  • Simba lab used a formula for the entire solution, so did they make up a formula or is there a list or something – user779444 Sep 30 '15 at 18:35
  • The limit he used was actually the derivative of $ln(x)$ evaluated at $x=1$, just use the limit definition of a derivative and you'll see what I mean. @user779444 – Ahmed S. Attaalla Sep 30 '15 at 20:59