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