$$\lim_{n\to +\infty}\ e^{\sqrt n }\cdot\Big(1 - \frac{1}{\sqrt n}\Big)^n$$
Answers are:
A)0
B)1
C)e
D)${\sqrt e}$
E)$\frac{1}{\sqrt e}$
I tried working on the second part to get it in a better form. In the end I got $e^{- \sqrt n}$. Returning at the beginning with this new form it would be:
$$\lim_{n\to +\infty}\ e^{\sqrt n }\cdot e^{- \sqrt n}$$ which would eventually turn into $$\lim_{n\to +\infty}\ e^{0}$$ so the answer would be B) 1 but it's not. The answer is E)$\frac{1}{\sqrt e}$ but I can't figure it out why.
Answer
Consider $$A= e^{\sqrt n } * \left(1 - \frac{1}{\sqrt n}\right)^n$$ Take logarithms of both sides $$\log(A)=\sqrt n+n\log\left(1 - \frac{1}{\sqrt n}\right)$$ Now, use the fact that for small values of $x$, $\log(1-x)=-x-\frac{x^2}{2}+O\left(x^3\right)$. Replace $x$ by $\frac{1}{\sqrt n}$ which makes $$\log(A)=\sqrt n+n(-\frac{1}{\sqrt n}-\frac{1}{2 n}+\cdots)$$ Expand and simplify.
I am sure that you can take from here.
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