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Show that \[\sqrt{x+2}\rightarrow2 \] as \[x \rightarrow \infty\]
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I'll save you the trouble, it doesn't.
\[\lim_{x \rightarrow \infty} \sqrt{x+2} = \sqrt{\lim_{x \rightarrow \infty} x+2} = \sqrt{\infty + 2} = \sqrt{\infty} = \infty\] I can apply the rule \[\lim_{x \rightarrow \infty} f(x) = f(\lim_{x \rightarrow \infty}x)\] because the function is continuous look at the function |dw:1346031956692:dw|
it will keep increasing to infinity look at the graph
Yeah sorry I got the question wrong.
no its ok, I hope my explanation was helpful
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