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

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Answer 1 · 2017-10-17T14:32:17

See below.

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 · 2017-10-17T14:47:04

#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"#