$\ \lim_{x\to0} \frac{x}{\cos(\frac{\pi}{2}-x)} $ without L'Hospital's rule

Question

How do I find the following?$$\ \lim_{x\to0} \frac{x}{\cos(\frac{\pi}{2}-x)} $$

I have tried to use trig identities :

$$ \frac{x}{\cos(\frac{\pi}{2}-x)} = \frac{x}{\cos\frac{\pi}{2}\cos x+\sin\frac{\pi}{2}\sin x} = $$

Can't really see anything out of that?

edit- so based on Math lover hint :

$$\ \frac{x}{\cos \frac{\pi}{2}\cos x + \sin \frac{\pi}{2}\sin x} = \frac{x}{0 \cdot \cos x + 1\cdot \sin x } = \frac{x}{\sin x} = \frac{0}{0} = 0 $$

@bm1125 Just use the last step as $$\lim_{x \to 0} \frac{1}{\frac{ \mathrm{sin}(x)}{x}}=1$$ — paulplusx, Sep 25 '18 at 20:20

score 6 · Accepted Answer · answered Sep 25 '18 at 19:59

6

hint

$$\cos(\frac \pi 2-x)=\sin(x)$$

answered Sep 25 '18 at 19:59

hamam_Abdallah

62,951

1

Great, then we only need to go down the following rabbit hole regarding sin(x)/x – String Sep 25 '18 at 20:05

score 1 · Answer 2 · answered Sep 25 '18 at 20:07

1

As an alternative by $f(x)=\cos\left(\frac{\pi}{2}-x\right)$ from the definition of derivative

$$\lim_{x\to0} \frac{x}{\cos\left(\frac{\pi}{2}-x\right)}=\lim_{x\to0} \frac{x-0}{\cos\left(\frac{\pi}{2}-x\right)-\cos\left(\frac{\pi}{2}\right)}=\frac1{f'(0)}=1$$

answered Sep 25 '18 at 20:07

user

154,566

Thanks but still don't have the knowledge of derivatives. – bm1125 Sep 25 '18 at 20:10
@bm1125 You are welcome! Indeed I gace that as an alternative, the direct method by $sin x/x \to 1$ is of course better. For the proof refer to sin x/x \to 1 – user Sep 25 '18 at 20:12

$\ \lim_{x\to0} \frac{x}{\cos(\frac{\pi}{2}-x)} $ without L'Hospital's rule

2 Answers2