I'm reviewing for my midterm in 3 hours and just came across this practice question/solution and don't understand it. Thank you!
2026-04-03 18:18:21.1775240301
Calculus, limits: Can someone explain to me why $-\frac{\sqrt{2+x^2}}{2x}$ is equal to $-\frac{1}{2}$ when $x$ approaches $\infty$?
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$$\lim\limits_{x \to \infty}-\dfrac{\sqrt{2+x^2}}{2x} = \lim\limits_{x \to \infty}-\dfrac{\sqrt{2+x^2}/x}{2x/x} = \lim\limits_{x \to \infty}-\dfrac{\sqrt{2+x^2}/\sqrt{x^2}}{2} = \lim\limits_{x \to \infty}-\dfrac{\sqrt{2/x^2+1}}{2}$$ $$= -\dfrac{\sqrt{0+1}}{2} = -\dfrac{1}{2}$$