Question

How do you find the limit of `(cos(x)/sin(x) + 1)` as x approaches 0 using l'hospital's rule?

Answer 1

See below.

Explanation:

Plugging in zero gives:

`1/0 + 1`

L'Hospital's Rule can only be used when we have a quotient of an indeterminate form:

`0/0` , `oo/oo`

Since we do not have this form, l'hospital's rule cannot be used.

If you evaluate the limit by plugging in values approaching 0 from the left the limit will be:

`lim_(x->0^-)cos(x)/sin(x) +1= -oo`

And from the right:

`lim_(x->0^+)cos(x)/sin(x) +1= oo`

So the limit is undefined

Using L'Hospital's rule:

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

`d/dx sin(x) = cos(x)`

`lim_(x->0)((-sin(x))/cos(x) + 1)= ((-sin(0))/cos(0)) +1 =( 0/1 + 1)=color(red)(1)`

`color(red)(lim_(x->0^)cos(x)/sin(x) +1 != 1)`

So L'Hospital's rule fails.

Answer 2

`lim_(x->0) (frac{cosx}{sinx}+1)`

`= lim_(x->0) frac{cosx+sinx}{sinx}`

By direct substitution, this gives `1/0`. This is not one of the indeterminate forms of L'Hospital's rule.

`lim_(x->0^-) frac{cosx+sinx}{sinx} = -oo`

`lim_(x->0^+) frac{cosx+sinx}{sinx} = oo`

`:. lim_(x->0) frac{cosx+sinx}{sinx} " DNE"`