Question

lim x approaches pi/2+ cosx/1-sinx use L'Hospitals rule

Answer 1

\[\lim_{x \rightarrow \pi/2^+}\frac{ cosx }{ 1-sinx }\]

Answer 2

yup

Answer 3

so this gives us the indeterminate form 0/0

Answer 4

yes

Answer 5

just don't understand y the answer equals - infinity

Answer 6

so we take the derivative with respect to x of the numerator and then of the denominator

Answer 7

i did

Answer 8

i got \[\lim_{x \rightarrow \pi/2+} tanx\]

Answer 9

i just don't understand y that equals \[-\]

Answer 10

- infinity

Answer 11

you mean the original limit? have you tried graphing it?

Answer 12

i answer for this problem is - infinity but when u plug in pi/2+ for x in tanx don't u use the unit circle so tan pi/2 is undefined right

Answer 13

y is it -infinity

Answer 14

yes pi/2 is undefined because as this function approaches pi/2 from the right, it goes to negative infinity