Boilanubjaini8f

2022-06-28

We must to calculate
$\underset{x\to \pi /2}{lim}\frac{\mathrm{sin}x-1}{2x-\pi }.$

Zayden Wiley

Expert

We must to calculate
$\underset{x\to \pi /2}{lim}\frac{\mathrm{sin}x-1}{2x-\pi }$
If we use a change of variables then this becomes:

where All we have done at (*) is use the definition of
But and therefore the answer is
Also, from the comments, you went wrong here:
$\underset{t\to 0}{lim}\frac{\mathrm{sin}\left(t+\frac{\pi }{2}\right)-\mathrm{sin}\frac{\pi }{2}}{2t}=\underset{t\to 0}{lim}\frac{2\mathrm{sin}\left(t+\pi \right)\mathrm{cos}t}{2t}$
This is incorrect. The correct formula is:
which would get you:

Yesenia Sherman

Expert

Another way to rewrite it is
$\frac{\mathrm{sin}\left(t+\pi /2\right)-1}{2t}=\frac{\mathrm{sin}t\mathrm{cos}\left(\pi /2\right)+\mathrm{sin}\left(\pi /2\right)\mathrm{cos}t-1}{2t}=\frac{\mathrm{cos}t-1}{2t}.$

Do you have a similar question?