How do you find the limit of #tantheta/theta# as #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, #0/0#, L'Hôpital's rule applies.

Take the derivative of the numerator

#(d[tan(theta)])/(d theta) = sec^2(theta)#

Take the derivative of the denominator

#(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#