Limit of Trig Function | Solving Without L'Hospital

[juantheron]

\lim_{x\rightarrow \frac{\pi}{2}}\frac{\sin \left(x\cos x\right)}{\cos \left(x \sin x\right)}

My Try:: \lim_{x\rightarrow \frac{\pi}{2}} \frac{\sin \left(x\cos x\right)}{x\cos x}.\frac{\frac{\pi}{2} - x\sin x}{\sin \left(\frac{\pi}{2} - x \sin x\right)}.\frac{x \cos x}{\frac{\pi}{2}-x \sin x}

Now Using The limit \lim_{y \rightarrow 0}\frac{\sin y}{y} = \lim_{y \rightarrow 0}\frac{y}{\sin y} = 1

So our Limit is Convert into \lim_{x \rightarrow \frac{\pi}{2}}\frac{x \cos x}{\frac{\pi}{2}-x \sin x}

Now Without Using L. Hospital , How Can I solve after That,

please Help me

Thanks

 

[Saitama]

$\cos(x)=\sin(\pi/2+x)$

 

[haruspex]

$\cos(x)=\sin(\pi/2+x)$

You mean $\cos(x)=\sin(\pi/2-x)$

 

[Saitama]

You mean $\cos(x)=\sin(\pi/2-x)$

No. Is it wrong to say $\cos(x)=\sin(\pi/2+x)$? I don't see anything wrong here.

 

[verty]

No. Is it wrong to say $\cos(x)=\sin(\pi/2+x)$? I don't see anything wrong here.

[strike]Cosine is negative in the second quadrant.[/strike] (sorry, I also didn't see it)

Juantheron: Using the linear approximations, cos(x) ≈ -x, sin(x) ≈ 0. I think this is easiest (I don't know how else to do it).

 

[vanhees71]

You should expand the numerator and denominator each around x=\pi/2. Your simplification is correct. So you can apply this idea to your simpler limit.

This is, however, not really different from using de L'Hospital's rule. Are you explicitly forbidden to use it?

@verty: The correct expansion of sine and cos around 0 is
\sin x = x-\frac{x^3}{3!} +\mathcal{O}(x^5), \quad \cos x=1-\frac{x^2}{2!}+\mathcal{O}(x^4),
but this doesn't help here.

 

[verty]

You should expand the numerator and denominator each around x=\pi/2. Your simplification is correct. So you can apply this idea to your simpler limit.

This is, however, not really different from using de L'Hospital's rule. Are you explicitly forbidden to use it?

@verty: The correct expansion of sine and cos around 0 is
\sin x = x-\frac{x^3}{3!} +\mathcal{O}(x^5), \quad \cos x=1-\frac{x^2}{2!}+\mathcal{O}(x^4),
but this doesn't help here.

Yes but cos at $π \over 2$ looks just like -sin at 0, and sin at $π \over 2$ looks just like -cos at 0. Knowing that sin and cos have the same shape is the proof. It is ad hoc but one could always do the Taylor expansion at $π \over 2$.

 

[Curious3141]

\lim_{x\rightarrow \frac{\pi}{2}}\frac{\sin \left(x\cos x\right)}{\cos \left(x \sin x\right)}

My Try:: \lim_{x\rightarrow \frac{\pi}{2}} \frac{\sin \left(x\cos x\right)}{x\cos x}.\frac{\frac{\pi}{2} - x\sin x}{\sin \left(\frac{\pi}{2} - x \sin x\right)}.\frac{x \cos x}{\frac{\pi}{2}-x \sin x}

Now Using The limit \lim_{y \rightarrow 0}\frac{\sin y}{y} = \lim_{y \rightarrow 0}\frac{y}{\sin y} = 1

So our Limit is Convert into \lim_{x \rightarrow \frac{\pi}{2}}\frac{x \cos x}{\frac{\pi}{2}-x \sin x}

Now Without Using L. Hospital , How Can I solve after That,

please Help me

Thanks

L' Hopital's is the most direct approach, of course. But if you can't use it, try putting converting every cosine to sine using $\displaystyle \cos y = \sin(\frac{\pi}{2} - y)$. You can employ this whenever the resulting sine argument becomes zero at the limit. The reason you do this is because sine has an easy approximation around zero. You then employ $\displaystyle \sin z$ ~ $\displaystyle z$ when $\displaystyle z \to 0$. This approach is even quicker than L' Hopital's.

I'll start you off.

$\displaystyle \frac{\sin(x\cos x)}{\cos(x\sin x)}$~ $\displaystyle \frac{x\cos x}{\frac{\pi}{2} - x\sin x}$

Can you proceed from there?

 

[Saitama]

I'll start you off.

$\displaystyle \frac{\sin(x\cos x)}{\cos(x\sin x)}$~ $\displaystyle \frac{x\cos x}{\frac{\pi}{2} - x\sin x}$

OP already reached that. He was stuck here so I suggested to use $\cos(x)=\sin(\pi/2+x)$ instead of $\cos(x)=\sin(\pi/2-x)$. That gives the answer in a few steps.

 

[Curious3141]

OP already reached that. He was stuck here so I suggested to use $\cos(x)=\sin(\pi/2+x)$ instead of $\cos(x)=\sin(\pi/2-x)$. That gives the answer in a few steps.

You don't have to change tack midstream here. Just apply the same thing iteratively while simplifying $\sin x$ ~ $1$.

I've edited out what I posted earlier as it is a complete solution. The OP should easily be able to get it from this point.

 

[vanhees71]

I don't know, why you don't just follow the idea to do the expansion of numerator and denominator around x=\pi/2.

Since \cos(\pi/2)=0 you have
\sin(x \cos x)=x \cos x + \mathcal{O}(x^2 \cos^2 x).
Further you can use
\cos x=-(x-\pi/2) + \mathcal{O}[(x-\pi/2)^3],
and for the denominator
\sin x=1+\mathcal{O}[(x-\pi/2)^2].
Now plugging all this in your limit
\lim_{x\rightarrow \frac{\pi}{2}}\frac{\sin \left(x\cos x\right)}{\cos \left(x \sin x\right)}
it should be straight forward to find this limit!

Of course, using de L'Hospital is much less effort after all, but if this is forbidden to use, what can we do...

You can also set x=\pi/2-x' and take the limit x' \rightarrow 0. This has the advantage that you can use
\sin x=\sin(\pi/2-x')=\cos x'=1+\mathcal{O}(x'^2), \quad \cos x=\cos(\pi/2-x')=\sin x'=x'+\mathcal{O}(x'^3).