Recent content by Nipuna Weerasekara

This is not a case whether you are allowed to use the L'Hopital law or not... It is a challenge where you cannot use it. Hence the question gets tricky... The following shows the method to solve the limit without using the law. ##\lim_{x\to0}\space \frac {sin(\pi (Cos ^2 (x)))}{x^2}##...

The answer is ##\pi##

Found the answer! Just an easy fix... ##\;\sin(\pi\cos^2x)=\sin(\pi-\pi\sin^2x)=\sin(\pi\sin^2x)##

Actually the real question here is to solve the limits question ##\lim_{x\to0}\space \frac {sin(\pi (Cos ^2 (x)))}{x^2}## without using L'Hopital's law...

Homework Statement Solve the following limit. $$\lim_{x\to0} \space \frac {sin(\pi (Cos ^2 (x)))}{\pi (Cos ^2 (x))}$$ The Attempt at a Solution When I plug ##x\to 0 ## into the limit, I get 0/1... Then what can I do? See here I can't even apply L'Hopital's law... Please help! Here I see the...

Thank you...

Yup you're correct, but then how do I do it. I have no idea how to get something like ##x^2\geq1## from ## x>-1##.

Homework Statement Let ##x\in \mathbb{R} ## Prove the conditional statement that, if ## x>-1## then ## x^2 + \frac {1}{x^2+1} \geq 1## 2. The attempt at a solution Suppose ## x>-1## is true. Then ## x^2>1## Then ## \frac{1}{2}>\frac {1}{x^2+1}## Then ##x^2+ \frac{1}{2}>x^2+\frac...

No I was wrong all along... I solved it. The statement is true. All r is real and greater than Golden Ratio (1.61803)...

But in a sense I can say that r is not strictly a real number hence the statement is false... Am I right?

I don't get what you say here.

Eventually one inequality leads that r has a complex variance. (by solving 3r^2 +3 > 3r)

Actually this gives me an idea how to approach the question. i can try to get a result for r by solving three inequalities $$3r^2 +3 > 3r$$ $$3r^2+3r>3$$ $$3+3r>3r^2$$

Actually there is a theorem called triangle inequality theorem, which states given ABC triangle, which has a,b,c length sides a+b>c , a+c>b and b+c>a must be true. therefore no need to show it to you that I can't create a triangle out of those sides 1,1 and 10.

I guess so