How do you find the limit of $\frac{tan θ}{θ}$ $tantheta/theta$ as $θ \to 0$ $theta->0$ ?

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Answer 1 · 2016-11-02T03:55:39

L'Hôpital's rule applies. Please see the explanation.

Because the expression evaluated at the limit is an indeterminate form, $\frac{0}{0}$ $0/0$ , L'Hôpital's rule applies.

Take the derivative of the numerator

$\frac{d [tan (θ)]}{d θ} = {sec}^{2} (θ)$ $(d[tan(theta)])/(d theta) = sec^2(theta)$

Take the derivative of the denominator

$\frac{d θ}{d θ} = 1$ $(d theta)/(d theta) = 1$

Assemble this into a fraction with the same limit:

$lim_(thetato0)(sec^2(theta))/1 = sec^2(0)= 1$

According to L'Hôpital's rule, the original expression goes to the same limit:

$lim_(thetato0)(tan(theta))/theta = 1$

How do you find the limit of tantheta/thetatanθθtantheta/theta as theta->0θ→0theta->0?