Taking the Limit of this fraction involving trig functions

[Memo]

Homework Statement

Lim (x->0) (cosx-sqrt(cos2x))/tan^2(x)
x∈(-π/4;π/4)\{0}

Relevant Equations

lim (x->x0) (1-cosa*f(x))/[f(x)]^2=a^2/2

Can't attempt to solve the task.
I'd appreciate it a lot if you could help!

 

[pasmith]

You can express <br /> f(x) = \frac{\cos x - \sqrt{\cos 2x}}{\tan^2 x} in terms of \cos x using basic trig identities. What can you do with <br /> \lim_{x \to 0} f(x) = \lim_{x \to 0} g(\cos x)?

 

[Memo]

lim (x->0) g * lim(x->0) cosx

 

[Orodruin]

lim (x->0) g * lim(x->0) cosx

You are misunderstanding #2. This is not how limits of a function of a function works.

 

[PeroK]

lim (x->0) g * lim(x->0) cosx

I think you should get the limit into a form where it is explicitly a function of $\cos x$ and take it from there.

 

[Memo]

this is what my teacher gave me. Yes, I don't understand him/her. Not sure if he/she's telling me to convert the function into that form or what, I don't see how it would turn out and just answered within my knowledge

 

[PeroK]

View attachment 334560
this is what my teacher gave me. Yes, I don't understand him/her. Not sure if he/she's telling me to convert the function into that form or what, I don't see how it would turn out and just answered within my knowledge

I think you should get the limit into a form where it is explicitly a function of $\cos x$ and take it from there.

 

[Memo]

I was replying to #4. This is as far as I can get to, sorry.

 

[PeroK]

I was replying to #4. This is as far as I can get to, sorry.
View attachment 334562

I can't see what you have on the right hand side. That doesn't look as helpful as I'd hoped.

Do you know l'hospital's rule?

 

[pasmith]

Write it as <br /> (\cos^2 x)\frac{\cos x - \sqrt{2\cos^2 x - 1}}{1 - \cos^2 x} = u^2 \frac{u - \sqrt{2u^2 - 1}}{1 - u^2}

 

[Memo]

Write it as <br /> (\cos^2 x)\frac{\cos x - \sqrt{2\cos^2 x - 1}}{1 - \cos^2 x} = u^2 \frac{u - \sqrt{2u^2 - 1}}{1 - u^2}

Could you tell me what happens after that?

 

[Orodruin]

You can express <br /> f(x) = \frac{\cos x - \sqrt{\cos 2x}}{\tan^2 x} in terms of \cos x using basic trig identities. What can you do with <br /> \lim_{x \to 0} f(x) = \lim_{x \to 0} g(\cos x)?

While a tractable path, I’d say it would be slightly more transparent to factorize out cos(x) and write the rest in terms of tan(x). This also directly uses the result regarding the product of limits the teacher apparently referred to.

 

[WWGD]

You can also use the limit definition of continuity. If g is continuous at $x_0$, then, $lim_{x\rightarrow x_0}g(x)=g(lim_{x\rightarrow x_0})$

 

[PeroK]

Here's another idea. Multiply top and bottom by the conjugate of numerator to get rid of the square root.

PS This works nicely!

 

[SammyS]

I was replying to #4. This is as far as I can get to, sorry.
View attachment 334562

Use the expression in the above image (post #8) - well, most importantly use the numerator.

$\displaystyle \dfrac{ \cos x-\sqrt{ 2\cos^2x -1 \, } } {\tan^2x}$

Then do as @PeroK suggested in the preceding post.