How to find the limit of this type of function
- Thread starter alexmay
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D
Deleted member 4993
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Do you know in which quadrants tan(x) is positive?
Do you know in which quadrants tan(x) is negative?
D
Deleted member 4993
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Correct -
Can you plot y = tan(x) [suggest use wolframalpha] and "see"
where does tan(x) "go" as you approach x = \(\displaystyle \frac{\pi}{2}\) from x = 0?
where does tan(x) "go" as you approach x = \(\displaystyle \pi\) from x = \(\displaystyle \frac{\pi}{2}\)?
D
Deleted member 4993
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I asked two questions in response 4.
Which are you answering in response 5?
Please answer both questions from response 4.
D
Deleted member 4993
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Did you plot the function - as I had suggested?
Try this
[MATH]\dfrac{\pi}{2} > h > 0, \text { and } x = \dfrac{\pi}{2} \implies 0 < sin(h) < 1,\ 0 < cos(h), \text { and }[/MATH]
[MATH]tan(x + h) = \dfrac{sin \left ( \dfrac{\pi}{2} + h \right )}{cos \left ( \dfrac{\pi}{2} + h \right )} = \dfrac{sin ( \pi/2) * cos(h) + sin(h) * cos(\pi/2)}{cos(\pi/2) * cos(h) - sin(\pi/2) * sin(h)} \implies[/MATH]
[MATH]tan(x + h) = \dfrac{1 * cos(h) + sin(h) * 0}{0 * cos(h) - 1 * sin(h)} = -\ \dfrac{cos(h)}{sin(h)} < 0.[/MATH]
[MATH]\therefore \lim_{x \rightarrow (\pi/2)^+} tan(x) = \lim_{h \rightarrow 0^+} \left (-\ \dfrac{cos(h)}{sin(h)} \right) = WHAT?[/MATH]
[MATH]\dfrac{\pi}{2} > h > 0, \text { and } x = \dfrac{\pi}{2} \implies 0 < sin(h) < 1,\ 0 < cos(h), \text { and }[/MATH]
[MATH]tan(x + h) = \dfrac{sin \left ( \dfrac{\pi}{2} + h \right )}{cos \left ( \dfrac{\pi}{2} + h \right )} = \dfrac{sin ( \pi/2) * cos(h) + sin(h) * cos(\pi/2)}{cos(\pi/2) * cos(h) - sin(\pi/2) * sin(h)} \implies[/MATH]
[MATH]tan(x + h) = \dfrac{1 * cos(h) + sin(h) * 0}{0 * cos(h) - 1 * sin(h)} = -\ \dfrac{cos(h)}{sin(h)} < 0.[/MATH]
[MATH]\therefore \lim_{x \rightarrow (\pi/2)^+} tan(x) = \lim_{h \rightarrow 0^+} \left (-\ \dfrac{cos(h)}{sin(h)} \right) = WHAT?[/MATH]
Steven G
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Graphing, whether it is in your head or on a piece of paper (computer screen, etc) is the easiest way to see this limit.
Studying calculus requires you to know algebra/precalculus and trigonometry. As a result you should know how to graph the tangent function. If not, it is no big deal but you better learn it before continuing with calculus.
Studying calculus requires you to know algebra/precalculus and trigonometry. As a result you should know how to graph the tangent function. If not, it is no big deal but you better learn it before continuing with calculus.
Jomo
I agree, but graphing does not explain why something is so. I suspect that is what post 9 is about. I further suspect that what was really confusing the student was a failure to grasp fully the concept of the limit from the right.
To put it a different way, graphing gives you answers rather than reasons. If what the student sought were reasons, then he would have to do the actual trigonometry.