# How do you find the limit of sin((x-1)/(2+x^2)) as x approaches oo?

Apr 20, 2016

Factorise the maximum power of $x$ and cancel the common factors of the nominator and denumerator. Answer is:

${\lim}_{x \to \infty} \sin \left(\frac{x - 1}{2 + {x}^{2}}\right) = 0$

#### Explanation:

${\lim}_{x \to \infty} \sin \left(\frac{x - 1}{2 + {x}^{2}}\right)$

${\lim}_{x \to \infty} \sin \left(\frac{1 \cdot x - 1 \cdot \frac{x}{x}}{2 \cdot {x}^{2} / {x}^{2} + 1 \cdot {x}^{2}}\right)$

${\lim}_{x \to \infty} \sin \left(\frac{x \cdot \left(1 - \frac{1}{x}\right)}{{x}^{2} \cdot \left(\frac{2}{x} ^ 2 + 1\right)}\right)$

${\lim}_{x \to \infty} \sin \left(\frac{\cancel{x} \left(1 - \frac{1}{x}\right)}{{x}^{\cancel{2}} \left(\frac{2}{x} ^ 2 + 1\right)}\right)$

${\lim}_{x \to \infty} \sin \left(\frac{1 - \frac{1}{x}}{x \left(\frac{2}{x} ^ 2 + 1\right)}\right)$

Now you can finally take the limit, noting that $\frac{1}{\infty} = 0$:

$\sin \left(\frac{1 - 0}{\infty \cdot \left(0 + 1\right)}\right)$

$\sin \left(\frac{1}{\infty}\right)$

$\sin 0$

$0$