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How to find this limit?

For $|f(x)-3|\le x^2$, $$\lim_{x\to0}\frac{f(x)-\sqrt{x+9}}{x}$$

Can I make $f(x)=x^2+3$, and then $$\lim_{x\to0}\frac{x^2+3-\sqrt{x+9}}{x}$$ Using l'Hopital's, $$\lim_{x\to0}\frac{2x-\frac{1}{2\sqrt{x+9}}}{1} = -\frac{1}{6}$$

I'm not confident about this answer. How do I do this problem?

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    "Can I make $f(x)=x^2+3$" No, that's not correct. – 2017-02-15
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    However, you can try writing $f(x) = f(x)+0 = f(x)-3+3$ – 2017-02-15
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    Simple but slightly tricky question +1 – 2017-02-15

1 Answers 1

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Note that $ -x^2 \leq (f(x) - 3) \leq x^2$, implying that $ 3-x^2 \leq f(x) \leq 3+x^2$ for all $x$. Hence, fixing $x$, we have that: $$ \frac{3-x^2-\sqrt{x+9}}{x} \leq \frac{f(x) - \sqrt{x+9}}{x} \leq \frac{3+x^2 - \sqrt{x+9}}{x} $$

Now, take the limit as $x \to 0$ on both sides of this inequality. Both the left and right sides converge to $\frac {-1}6$ as $x \to 0$ (you can do l'hopital for this).

Hence, the answer is $\frac{-1}{6}$ by the squeeze theorem, but not because you assumed $f$ to be a certain function, only using what was given in a clever manner.

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    you was faster than me +1. – 2017-02-15
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    @haqnatural Thank you for the upvote! If you honestly think that your answer has a different idea, do not delete it. I will upvote it if I find it interesting (and the questioner). – 2017-02-15
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    you are welcome,your answer is more detailed of course) – 2017-02-15