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

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Answer 1 · 2018-04-11T00:38:46

$-1/2$

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

$L = -1/2$

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