Limits to solve for equation of line

Asked Oct 24 '16 at 12:35

Active Oct 25 '16 at 01:59

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If $\lim\limits_{x \to 0^+} f(x)=3$, $\lim\limits_{x \to 0^-}\ f(x)=3$ and $\ f(0)=5,f'(x)<0$ for x belonging to (-inf,0) and $\ f'(x)>0$ for x belonging to (0,Inf) , then the LINE $\lim\limits_{x \to 0}\ f(\cos^3x-\cos^2x) = x \lim\limits_{x \to 0}\ f(\sin^2x-\sin^3x)$ is ?

PS, can that x on RHS be taken inside the limit, even if that wouldnt help in this case?

edited Oct 25 '16 at 01:59

asked Oct 24 '16 at 12:35

Shashank Holla

HINT: $lim_{x\to0}(\cos^3x-\cos^2x)=0$ and $lim_{x\to0}(\sin^2x-\sin^3x)=0$. – Intelligenti pauca Oct 24 '16 at 14:49
But that would make the equation 0. The equation is supposed to be of a line – Shashank Holla Oct 25 '16 at 01:50
The line equation turns out to be $3=3x$. See here: http://math.stackexchange.com/questions/456029/change-of-variables-in-limits – Intelligenti pauca Oct 25 '16 at 19:44
since cos^3x-cos^2x is less that zero always, the function on LHS tends to zero only from left. And the function on right tends to zero only from right. So the limit x->0^- must be for LHS and 0^+ for the function on RHS. Is that right? – Shashank Holla Oct 26 '16 at 11:10
Yes that's right, but both limits are equal to 3, so in practice that is not so relevant. – Intelligenti pauca Oct 26 '16 at 11:21

Limits to solve for equation of line

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