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I'm trying to solve the limit of this inequality. The question goes as follows: If $$4x - 9 \leq f(x) \leq x^2 - 4x + 7$$ for $x \geq 0$, find $\lim_{x\to 4} f(x)$.

I'm not really sure how to go about this problem. I tried solving it using one-sided limits and got my answer as 7 but I'm not sure how to elegantly present my answer. I want to know how to present my answer in an appropriate mathematical way.

But I'm not sure if there is a way to solve it using Sandwich theorem. I would like to know if it's possible, as well.

Thanks.

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  • $\begingroup$ Formally: I assume you see how to formalize that $\lim_{x\to 4}4x-9=7$ and that $\lim_{x\to 4}x^2-4x+7=7$. Using this, proceed as follows: For any $\epsilon>0$ there are positive $\delta_1$ and $\delta_2$ such that if $0<|x-4|<\delta_1$, then $-\epsilon<(4x-9)-7$, and if $0<|x-4|<\delta_2$, then $(x^2-4x+7)-7<\epsilon$. Let now $\delta=\min\{4,\delta_1,\delta_2\}$, and verify from the given inequality that if $0<|x-4|<\delta$, then $-\epsilon<f(x)-7<\epsilon$. $\endgroup$ – Andrés E. Caicedo Apr 25 '14 at 20:51
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Using the Sandwich Theorem, we have $\lim_{x \to 4} 4x-9 = 7$ and $\lim_{x \to 4} x^2-4x+7 = 7$, and since $f(x)$ is sandwiched in between, $\lim_{x \to 4} f(x) = 7$.

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I don't know if this is an accepted way of answering, but here is my solution: http://i.imgur.com/YggwNxO.png

Using the sandwich theorem, we find that f(x) is "sandwiched" between the two functions. Ergo, where the two functions meet, f(x) has nowhere to go but the same point.

Try drawing it out and see what it tells you. I think you'll find it very intuitive.

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In simplest terms, x approaches 4. So plug 4 in for x in both sides. You get 7 <= f(x) <= 7 By simple logic, you can determine there are no values both smaller than 7 and greater than 7, so f(x) must be 7.

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