Find $\lim_{x \to \infty} (-1)^{x-1}\sin(\pi\sqrt{x^2+0.5x+1})$ with $x\in\mathbb{N}$ 
Evaluate
  $$\lim_{x \to \infty} (-1)^{x-1}\sin(\pi\sqrt{x^2+0.5x+1})$$
  where $x$ belongs to the natural numbers.

Guys, this is what I have done:
I first took $x$ out of root and in this process I took $0.5/x$ and $1/x$ as zero, so now I have $\sin(\pi x)$.
We know that $\sin(\pi x)=0$ for $x$ belonging to natural numbers
so answer must be zero, but the answer is given as $-\frac{1}{\sqrt{2}}$.
I don't know what's wrong in my procedure.
 A: Note that 
$$\sqrt{x^2+0.5x+1}-x=\frac{0.5x+1}{\sqrt{x^2+0.5x+1}+x}$$
Then, if $x$ is an integer, recalling that $\sin(t-x\pi)=(-1)^x\sin(t)$, it follows that
$$(-1)^{x-1}\sin(\pi\sqrt{x^2+0.5x+1})=-\sin(\pi\sqrt{x^2+0.5x+1}-\pi x)\\=
 - \sin\left(\frac{\pi(0.5x+1)}{\sqrt{x^2+0.5x+1}+x}\right).$$
Can you take it from here?
A: This question can be solved in another way too.
We see that we have $\sqrt(x^2+0.5x+1)$ in the argument of sin function. Now clearly the given quadratic equation is positive for any value of 'x' because the D(determinant) value is always negative. So for any value of positive x this equation doesn't have any real solution. Now also we know that the value of sinx can never be greater than 1. So the argument inside the sin value is a very large positive value(as x tends to infinity) which gives the sin() value equal to 1(i.e as $x\to \infty sin(\pi\sqrt(x^2+0.5x+1))$ will tend to 1).Now the entire limiting value depend on $(-1)^(x-1)$ or $-(-1)^x$. This is a type of indeterminant form so this can be solved using L'Hospital's rule.
