What is the limit of the given inequality

Question

Suppose the inequality $\frac {1}{2}-\frac{x^2}{24}<\frac{1-\cos(x)}{x^2}<\frac {1}{2}$ then $\lim_{x\to 0}\frac{1-\cos(x)}{x^2}$ is? I already solved this question by taking limit on the inequalities the answer is $\frac{1}{2}<\lim_{x\to0}\frac{1-\cos(x)}{x^2}<\frac{1}{2}$ but I have a small doubt, as we know if $a<f(x)<a \implies f(x)=\phi$ so the solution is $1/2$ or doesn't exist?

Answer 1

For all $x\ne 0$,

$$\left(\frac12-\frac{x^2}{24},\frac12\right)\ne \emptyset$$ and $\dfrac12$ is an accumulation point.

Limits are always computed in neighborhoods, not on the point itself.

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You may reassure yourself by writing

$$\frac12-\frac{x^2}{24}\le f(x)\le\frac12$$ so that the intervals are closed, but the conclusion remains the same.

Answer 2

Recall that by squeeze theorem from

$$\frac {1}{2}-\frac{x^2}{24}<\frac{1-\cos(x)}{x^2}<\frac {1}{2}$$

since

$$\frac{1}{2}-\frac{x^2}{24}\to \frac12$$

we can conclude that

$$\frac{1-\cos(x)}{x^2}\to \frac12$$

Answer 3

Limits don't necessarily preserve strict inequalities. If you know that $f(x) < g(x)$ and $f(x)\to A$ and $g(x) \to B$, then you can only conclude $A \le B$ (not $A < B$).