$\begingroup$
$\endgroup$
2
I am having trouble with this question:
if f(x) is a continous function and $$ 1 - x^2 \leq f(x) \leq e^x $$ for all values of x, calculate $$ \lim_{x\to0} f(x) $$.
Do I have to calculate the limit on both sides of $f(x)$? How would I got about solving this?
2 Answers
$\begingroup$
$\endgroup$
1
A function $f$ with the given property cannot exist. For $x=-\frac12$ we would have $$ 0.75 = 1-(-1/2)^2 \le f(x) \le e^{-1/2} \approx 0.61 $$ but $0.75 \not\le 0.61$.
$\begingroup$
$\endgroup$
1
Yes, take limit on all three sides: $$1=\lim_{x\to0}(1-x^2)\le\lim_{x\to0}f(x)\le\lim_{x\to0}e^x=1$$
