How do you use L'hospital's rule to find the limit #lim_(x->0)(x-sin(x))/(x-tan(x))# ?

1 Answer
Apr 11, 2018

#-1/2#

Explanation:

The limit

#L = lim_(x->0)(x-sin(x))/(x-tan(x))#

is of the form #0/0#, Thus, we can use L'hospital's rule, which says

If #f(a)=g(a) = 0#, # lim_{x to a} f(x)/g(x) = lim_{x to a} {f^'(x)}/{g^'(x)}#

Thus,

#lim_(x->0)(x-sin(x))/(x-tan(x)) = lim_(x->0){d/dx(x-sin(x))}/ {d/dx(x-tan(x))}#

#qquad = lim_(x->0)(1-cos(x))/(1-sec^2(x))#

This, again is of the #0/0# form, so we use L'hospital's rule again

#L = lim_(x->0){d/dx(1-cos(x))}/{d/dx(1-sec^2(x))} = lim_(x->0)(sin(x))/((-2sec^2(x)tan(x)))#

We could use L'hospital's rule yet again, but it is much simpler to use simple trigonometry to go ahead :

#L = lim_(x->0)(sin(x))/((-2sec^2(x)tan(x))) = lim_(x->0)(1)/((-2sec^3(x))#

and thus

#L = -1/2#