Question

lim x->0
((cos t) -1)/ sin t

Answer 1

\[\lim_{x \rightarrow 0} \frac{ \cos x -1 }{ \sin t }\]

Answer 2

try multiply top's conjugate on top and bottom

Answer 3

\[\lim_{x \rightarrow 0}\frac{\cos(x)-1}{\sin(x)} \cdot \frac{\cos(x)+1}{\cos(x)+1} \\ =\lim_{x \rightarrow 0}\frac{\cos^2(x)-1}{\sin(x)(\cos(x)+1)}\]
and what is 1-cos^2(x)=? (think trig identity)

Answer 4

just change to neg sin^2(x)

Answer 5

ok great you should see something cancel from top and bottom now

Answer 6

so neg sinx/(cos x +1)

Answer 7

so 0

Answer 8

\[\lim_{x \rightarrow 0}\frac{-\sin^2(x)}{\sin(x)(\cos(x)+1)}=\lim_{x \rightarrow 0}\frac{-\sin(x)}{\cos(x)+1}\]
yep :)

Answer 9

tyty

Answer 10

np :)