# Not sure how to find the limit of this inequality?

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.

• 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 $-\epsilon<f(x)-7<\epsilon$. – Andrés E. Caicedo Apr 25 '14 at 20:51

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, $\lim_{x \to 4} f(x) = 7$.