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I was presented with theorems which state: If $\lim_{x\to a}f(x) = l$ and $\lim_{x\to a}g(x) = m$, then $$\lim_{x\to a}(f \cdot g)(x) = l \cdot m$$ $$\lim_{x\to a}(f + g)(x) = l + m$$ if $m \neq 0$,$$\lim_{x\to a}\Bigl(\frac 1g\Bigr)(x) = \frac 1m$$

When solving for example$$\lim_{x\to \infty}\sqrt{1 + \frac 1x},$$ I've seen teachers take the limit of each value in the radicand separately. What allows them to initially evaluate the limit of the radicand in the first place?

Ethan Bolker
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Sourblob
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  • Simply the fact that the function $\sqrt{\cdot}$ is continuous hence $\lim\sqrt{h(x)}=\sqrt{\lim h(x)}$ if $\lim h(x)$ exists and is finite and nonnegative. – Did Jul 11 '17 at 10:15

2 Answers2

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Good quesion. In fact, you have to be fairly careful when doing this.

It's because if $f$ and $g$ are continuous functions, then $$\lim_{x\to a}{f(g(x))} = f(\lim_{x\to a} g(x)$$ In your case, you can take $f(x)=\sqrt x$ and get the solution.

5xum
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Continuity. The square root function is continuous. And the definition of continuity is that $$\lim_{x\to a}g (x)=g (a). $$ So, if $\lim_{x\to\infty}f (x)=a $, then $$\lim_{x\to\infty}g (f (x))=g (a). $$

Martin Argerami
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