Please, I've been looking at it for a half an hour but can't find a mistake. The limit is supposed to be 2/3, I have to find it using L'Hopital's law. I've only used the fact that lim (sinx/x)^2 is 1 at x=0. It gives wrong results, although seems to be so trivial. ?
Easy limit question
- Thread starter Darya
- Start date
Cubist
Senior Member
- Joined
- Oct 29, 2019
- Messages
- 1,702
You went wrong at this stage, when you eliminated the crossed out parts...
[math] \lim_{x\to0}{ \frac{1-\frac{\xcancel{x^2}\cos^2x}{\xcancel{sin^2x}} }{x^2} }[/math]
Try to stick to these properties of limits since you can't pick arbitrary sub-expressions when applying limits.
EDIT: I've never seen ctg(x) before meaning cotangent. I've always used cot. The things you discover on this forum ?
[math] \lim_{x\to0}{ \frac{1-\frac{\xcancel{x^2}\cos^2x}{\xcancel{sin^2x}} }{x^2} }[/math]
Try to stick to these properties of limits since you can't pick arbitrary sub-expressions when applying limits.
EDIT: I've never seen ctg(x) before meaning cotangent. I've always used cot. The things you discover on this forum ?
Cubist
Senior Member
- Joined
- Oct 29, 2019
- Messages
- 1,702
Let me explain further... when aiming to eliminate the above, you'd have to start by trying to isolate the x^2/sin^2(x) within it's own limit. In order to do this you'd start by separating the initial division like so...
[math] \lim_{x\to0}{ \frac{1-\frac{x^2\cos^2x}{sin^2x} }{x^2} } [/math]
[math] = \lim_{x\to0}{ \left( 1-\frac{x^2\cos^2x}{sin^2x}\right) } / \lim_{x\to0}{x^2} \,[/math] ...BUT THIS IS NOT OK, see below
...the properties of limits page states, "the identity for division requires that the denominator on the right-hand side is non-zero". Unfortunately we get accustomed to applying limits without explicitly showing the whole process involved (otherwise it would take a long time to write everything out). But you need to be aware of exactly what is happening when you do several steps in one.