use L"Hospitals Rule to solve lim x-->pi/2^- tanx+secx?
回答 (4)
2014-07-26 11:59 pm
lim x→pi/2- , tanx + secx = (sinx + 1) / cosx = (1+1) / 0+ = + ∞
We got (1+1) / 0+ = 2/0+ which is not the same with 0/0 or ∞/∞
=> you cannot apply L'Hospital Rule.
Hope this helps.
We got (1+1) / 0+ = 2/0+ which is not the same with 0/0 or ∞/∞
=> you cannot apply L'Hospital Rule.
Hope this helps.
2014-07-26 11:43 pm
lim x-->pi/2^- tanx+secx = infty
2014-07-27 12:25 am
lim(x → π/2-)tanx + secx = lim(x → π/2-)(sinx + 1)/cosx = 2/0, which is not of the form 0/0 or ∞/∞ (and cannot be put into one of those forms) so l'Hopital's rule does not apply. Since sinx + 1 > 0 in the vicinity of x = π/2, we need only look at the sign of cosx. The limit from the left is ∞, while the limit from the right is -∞ .
2014-07-27 12:40 am
what does the symbolism π/2^- mean ?...if the problem is lim { x ---> π / 2 } ( -tan x + sec x ) =..
then L'H can be used as you would really have [ - sin x + 1 ] / cos x---> 0/0...L'H would
then yield [ - cos x / 1 ] --->0
then L'H can be used as you would really have [ - sin x + 1 ] / cos x---> 0/0...L'H would
then yield [ - cos x / 1 ] --->0
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