<s>[INST] Limit with parameter $p$ Find $ p $ that will make limit convergent
$$ \lim _ { x \to \infty } \left( \left( n + \left( \frac { x ^ 2 + 2 } { x ^ 2 - x + 1 } \right) ^ { 2 x + 1 } \right) \cdot \sin \left( \frac { x \pi } 2 \right) \right) $$
I know how to solve most of the basic limits. In this problem I don't know if I'm allowed to separate it and make $ 3 $ or $ 4 $ small ones. I thought about making $ p \sin.. + \sin*(big bracket)$ and then separating again because $ p $ is parameter and can go in front of limit and I wanted to separate second one so I can find big limit (it's $e^2$ or something but that doesn't matter).
 [/INST] $$ \lim _ { x \to \infty } \left( \left( n + \left( \frac { x ^ 2 + 2 } { x ^ 2 - x + 1 } \right) ^ { 2 x + 1 } \right) \cdot \sin \left( \frac { x \pi } 2 \right) \right) $$
since as mentioned the SIN is diverging because it keeps oscillating. Therefore, the first part has to converge to 0.
We have that the first part converges to $(n+e^2)$ Therefore your n should be $-e^2$.
Detailing the Calculation
$$\lim\limits_{x\to\infty}\left(\frac{x^2+2}{x^2-x+1}\right)^{2x+1}$$
$$=\lim\limits_{x\to\infty}\exp\left(\ln\left(\left(\frac{x^2+2}{x^2-x+1}\right)^{2x+1}\right)\right)$$
$$=\exp\left(\lim\limits_{x\to\infty}\ln\left(\left(\frac{x^2+2}{x^2-x+1}\right)^{2x+1}\right)\right)$$
$$\left(\exp\left(\left(1+2x\right)\ln\left(\frac{1+\frac{2}{x^{2}}}{1-\frac{1}{x}+\frac{1}{x^{2}}}\right)\right)\right)$$
Here we have to use the definition of $ln(1+x)=x$ if x goes to 0 and subtract the numerator from the denominator after applying the small transformation.
$$\exp\left(\left(2x+1\right)\left(\frac{1}{x^{2}}+\frac{1}{x}\right)\right)$$
$$\exp\left(\frac{2x}{x^{2}}+\frac{2x}{x}\right)$$
$$exp(2)$$
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