Question

a) Does the sequence (3+(-1)ⁿ) converge to 2? Justify.

 

 b) Show that $\lim _{x\to \infty }\left(\frac{x-3}{x+1}\right)^x=e^{-4}$

c) Check whether the sequence f_n (x) = $\frac{3x}{1+nx^2}$ where x ∈ [2,∞ [ is uniformly 

convergent in [2,∞ [

Answer 1

1) The sequence 

$\left(3+\left(-1\right)^n\right)$

doesn't converge to 2, because this sequence is oscillating between 2 and 4. Hence, it is divergent.

2) $\lim_{x\rightarrow\infty}{\left(\frac{x-3}{x+1}\right)^x}=\lim_{x\rightarrow\infty}{\left(\frac{x+1-4}{x+1}\right)^x}=$ $\lim\limits_{x\rightarrow\infty}{\left(\left(1+\frac{-4}{x+1}\right)^\frac{x+1}{-4}\right)^{-\frac{4x}{x+1}}}=$

$=\lim_{x\rightarrow\infty}{e^{-\frac{4}{1+\frac{1}{x}}}}= e^{-4}$.

3) if x=2 then $f_n\left(2\right)=\frac{6}{1+4n}$ and $\lim_{n\rightarrow\infty}{f_n\left(2\right)}=\lim_{n\rightarrow\infty}{\frac{6}{1+4n}}=0$.

If x>2 then $\left|f_n\left(x\right)\right|=\left|\frac{3x}{1+nx^2}\right|\le\left|\frac{3x}{nx^2}\right|=\left|\frac{3}{nx}\right|$ and $\lim_{n\rightarrow\infty}{\frac{3}{nx}}=0$

Conclution:

$f_n\left(x\right)\rightrightarrows0,\ x\in\left[2,+\infty\right[$