Question
Tue March 20, 2012 By: Sweta Ghosh

# Evaluate the following:

Wed March 21, 2012
Let y = tanxcosx
Taking limit both sides we get
lim x ? (pi/2)? (tan x)^(cos x)
you have to rewrite it as:
lim x ? (pi/2)? e^ln [(tan x)^(cos x)], that is, according to log properties,
lim x ? (pi/2)? e^[(cos x) ln (tan x)] =
which you'd better rewrite as a ratio:
lim x ? (pi/2)? e^{[ln (tan x)] / [(1/cos x)] = (e^(?/?))
so that you can apply L'Hopital's at the exponent as:
lim x ? (pi/2)? e^{D[ln (tan x)] /D [(1/cos x)] =
lim x ? (pi/2)? e^{[(1/cosÂ²x) / tan x] / [-(-sinx)/cosÂ²x)] =
lim x ? (pi/2)? e^{[(1/cosÂ²x) / (sin x /cos x)] / [-(-sinx)/cosÂ²x)]} =
lim x ? (pi/2)? e^{[(1/cosÂ²x) (cos x / sin x)] / (sinx/cosÂ²x)} =
lim x ? (pi/2)? e^{[1/(sin x cosx)] / (sinx/cosÂ²x)} =
lim x ? (pi/2)? e^{[1/(sin x cosx)](cosÂ²x /sin x)} =
lim x ? (pi/2)? e^[cosÂ²x/(sinÂ²x cosx)] =
lim x ? (pi/2)? e^(cosx/sinÂ²x) = e^(0/1Â²) = e?= 1
Now the given expression is log[tancosx] = logy = log1 = 0
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