sqrt(n^2+n)-n
|cos(n*pi/2)+sin(n*pi/2|(2n-1)/(n+2)
i can kind of start the first one and can't start the second one. any ideas?
So do you not want help with the first one? I cannot understand what you mean at all in number 2. Please rewrite with more parentheses. I don't know what you mean by | |. Are those absolute value brackets or are they another form of grouping? Make a clear denominator and numerator, please. If there is more than one term being multiplied together, then use spacing between the terms to show that.
Is that what you mean? And are you referring to the sequences? In other words, $\displaystyle \lim_{n\to\infty}\sqrt{n^2+n}-n$ and $\displaystyle \lim_{n\to\infty}\frac{|\cos(\frac{n\pi}{2})+\sin( \frac{n\pi}{2})|(2n-1)}{n+2} $
The first one's a classic example of a convergent limit involving $\displaystyle \infty-\infty$. Try multiplying by $\displaystyle \frac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n}$ which gives you $\displaystyle \frac{n^2+n-n^2}{\sqrt{n^2+n}+n} = \frac{n}{\sqrt{n^2+n}+n}$. Can you take it from here?
Spoiler:
For the second one, break the limit up into $\displaystyle \left(\lim_{n\to\infty}\left|\cos\frac{n\pi}{2}+\s in\frac{n\pi}{2}\right|\right)\left(\lim_{n\to\inf ty}\frac{2n-1}{n+2}\right)$.
Calculate the limits separately. If one of them diverges, the whole thing does (unless one of them goes to zero, then you need to be careful). If both of them converge, the whole limit converges to the product of the two limits.
Spoiler:
sorry about the confusion on the way i wrote the problem. yea, you got it right redsoxfa325 on how the problem should be written.
for the first problem, i'm not really able to see what the next step would be.
and for the second problem, i got that it diverges. is that right??