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October 22, 2014

October 22, 2014

Posted by **Jon** on Sunday, November 11, 2007 at 12:55pm.

I need to show work, so formatting answers in this manner would be most appreciated. Thanks in advance! :) :)

- Calculus -
**Anonymous**, Sunday, November 11, 2007 at 1:28pmEvaluate the limit as h -> 0 of:

[tan (pi/6 + h) - tan(pi/6)]/h

I thought the answer was √3/3, or tan(pi/6, but apparently that is wrong, any tips here?

- Calculus -
**Anonymous**, Sunday, November 11, 2007 at 1:29pmSorry, I mean to make a new question, disregard above 'answer'.

- Calculus -
**Jon**, Sunday, November 11, 2007 at 1:43pmGee, now you're going to make people who open this topic think the answer has already been provided. Thanks a ton...

- Calculus -
**SuperGirl**, Monday, May 31, 2010 at 7:44pmYou're a jerk.

- Calculus -
**H H Chau**, Friday, August 22, 2014 at 7:10pmtan(θ)=x/1

cos(θ)=1/sqrt(x^2+1)

sec^2(θ)=x^2+1

dθ/dt=3 rev/min=6π rad/min

x=tan(θ)

dx/dt=sec^2(θ) dθ/dt

At x=0.5 km

dx/dt=(0.5^2+1)*6π=23.6 km/min=1414 kph

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