Help me please
I'm not sure if this limit is correct but
can u guys solve for me

-1

Step-by-step explanation:

To solve this without L'Hopital's rule, multiply by the conjugate, (1 − cos n + sin n) / (1 − cos n + sin n).

The numerator becomes:

(1 − cos n + sin n) (1 − cos n + sin n)

(1 − cos n + sin n) − cos n (1 − cos n + sin n) + sin n (1 − cos n + sin n)

1 − cos n + sin n − cos n + cos² n − sin n cos n + sin n − sin n cos n + sin² n

2 − 2 cos n + 2 sin n − 2 sin n cos n

2 − 2 cos n + sin n (2 − 2 cos n)

(1 + sin n) (2 − 2 cos n)

The denominator becomes:

(1 − cos n − sin n) (1 − cos n + sin n)

(1 − cos n)² − sin² n

1 − 2 cos n + cos² n − sin² n

1 − 2 cos n + cos² n − 1 + cos² n

2 cos² n − 2 cos n

-cos n (2 − 2 cos n)

Dividing, the 2 − 2 cos n factors cancel out, leaving us with:

(1 + sin n) / -cos n

The limit as n approaches 0 is therefore:

(1 + 0) / -1

-1

50-38=12

step-by-step explanation:

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°