Question #f03d1

3 Answers
Sakuya Izayoi
Oct 11, 2017

#0.5#

Explanation:

This is because sin of 0 is 0. (On the unit circle, when it's 0 degrees, the point is at 0.)

Tangent is #sin/cos# and even though #cos(0)=1# (because of the unit circle) #0/1=0#.

Since it doesn't directly touch zero, we can try plugging in something extremely small, such as 0.01.

On the top, sin of 0.01 is roughly 0.00999 or 0.01. On the bottom, is 0.01000033, or roughly 0.01. #0.01/(0.01+0.01)=0.5#. The values become much more rounded as the number that is plugged in becomes extremely small.

Douglas K.
Oct 11, 2017

#lim_(theta rarr 0) sin(theta)/(theta+tan(theta))= 1/2#

Explanation:

Given: #lim_(theta rarr 0) sin(theta)/(theta+tan(theta))#

Because the expression evaluated at 0 yields the indeterminate form, #0/0#, then one should use L'Hôpital's rule by differentiating the numerator with respect to x and the denominator with respect to x:

#lim_(theta rarr 0) ((d(sin(theta)))/dx)/((d(theta+tan(theta)))/dx)#

#lim_(theta rarr 0) cos(theta)/(1+sec^2(theta))#

This expression can be evaluated at the 0:

#cos(0)/(1+sec^2(0)) = 1/(1+1) = 1/2#

L'Hôpital's rule states that, as goes the above limit, so goes the original limit:

#lim_(theta rarr 0) sin(theta)/(theta+tan(theta))= 1/2#

Jim H
Oct 11, 2017

#sintheta/(theta+tantheta) = (sintheta/sintheta)/(theta/sintheta+sin theta/(costheta sin theta) #

# = 1/(theta/(sintheta)+1/costheta)#

Taking limit as #theta rarr0# yields:

# = 1/(1+1/1) = 1/2#

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