Can You Solve These Challenging Calculus Problems?

[l'Hôpital]

Actually all you need is your trig identities:

sinx / x = cosx ; and since cos (0) = 1 ; sin (0) / 0 = 1

What? Take x = pi.

$$\frac{sin \pi}{\pi} = 0$$

But...

$$cos \pi = -1$$

 

[Redbelly98]

Let t = time when this question was asked.

Find t.

 

[weagle2008]

What? Take x = pi.

$$\frac{sin \pi}{\pi} = 0$$

But...

$$cos \pi = -1$$

The question isn't as x approaches Pi it is as x approaches 0. As anyone knows Pi = 180 degrees. 2*Pi = 360 degrees which is = 0 in trig. Thus cos (Pi) = -1, but cos (0) = 1.

 

[Anonymous217]

The question isn't as x approaches Pi it is as x approaches 0. As anyone knows Pi = 180 degrees. 2*Pi = 360 degrees which is = 0 in trig. Thus cos (Pi) = -1, but cos (0) = 1.

But you stated that (sinx)/x = cosx, which means that this is true for all x (which is wrong). This is what he was talking about.

sinx / x = cosx ; and since cos (0) = 1 ; sin (0) / 0 = 1

I would conjecture that you meant $$\lim_{x\rightarrow 0}\frac{sinx}{x} = \lim_{x\rightarrow 0} cosx$$. However, I'm not exactly sure how you got to that without L'H.