How do you find the limit of cos⁡x1−sin⁡x as x approaches π2?

Question

How do you find the limit of cos⁡x1−sin⁡x  as x approaches  π2?

Heather Fulton · Accepted Answer

Explanation:
If you try to evaluate the limit at

\frac{π}{2}

you obtain the indeterminate form

\frac{0}{0}

this means that LHôpitals rule applies.
To implement the rule, take the derivative of the numerator:

\frac{d \cos (x)}{d x} = - \sin x

take the derivative of the denominator.

\frac{d (1 - \sin (x))}{d x} = - \cos (x)

Assemble this into a fraction:

lim_{x \Rightarrow \frac{x}{2}} \frac{- \sin (x)}{- \cos (x)}

Please observe that the above is the tangent function:

lim_{x \Rightarrow \frac{x}{2}} \tan (x)

It is well known that the tangent function approaches infinity as x approaches

\frac{π}{2}

, therefore, the original expression does the same thing.

Melinda McCombs · Accepted Answer

Multiply by 1+sin⁡x1+sin⁡x  Explanation:  cos⁡x(1−sin⁡x)⋅(1+sin⁡x)(1+sin⁡x)=1+sin⁡xcos⁡x  Now as x⇒(π2)+ we have  1+sin⁡x⇒2 and  cos⁡x⇒0  So  limx⇒(x2)+cos⁡x1−sin⁡x=limx⇒(x2)+1+sin⁡xcos⁡x=−∞

RizerMix · Accepted Answer

I did my homework much faster with your help, the best, thanks for the help

How do you find the limit of \frac{\cos x}{1-\sin x}\text{

Answered question

Answer & Explanation

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