Question

How do you find the limit of `cosx/(1-sinx)` as x approaches pi/2+?

Answer 1

Use L'Hôpital's rule to discover that it approaches infinity as x approaches `pi/2`

Explanation:

If you try to evaluate the limit at `pi/2` you obtain the indeterminate form `0/0`; this means that L'Hôpital's rule applies.

To implement the rule, take the derivative of the numerator:

`(d{cos(x)})/dx = -sin(x)`

take the derivative of the denominator.

`(d{1 - sin(x)})/dx = -cos(x)`

Assemble this into a fraction:

`lim_(x->pi/2) (-sin(x))/(-cos(x))`

Please observe that the above is the tangent function:

`lim_(x->pi/2) tan(x)`

It is well known that the tangent function approaches infinity as x approaches `pi/2`, therefore, the original expression does the same thing.

Answer 2

Multiply by `(1+sinx)/(1+sinx)`

Explanation:

`cosx/((1-sinx)) * ((1+sinx))/((1+sinx)) = (1+sinx)/cosx`

Now as `xrarr(pi/2)^+`, we have

`1+sinx rarr 2` and

`cosx rarr 0^-`

So

`lim_(xrarr (pi/2)^+) cosx/(1-sinx) = lim_(xrarr(pi/2)^+) (1+sinx)/cosx = -oo`