Question #7a84b

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Answer 1 · 2017-03-27T12:27:26

#lim_(x->0) (1/sinx-1/tanx) = 0#

We have to evaluate the limit:

#lim_(x->0) (1/sinx-1/tanx)#

use the trigonometric identity: #tanx = sinx/cosx#

#lim_(x->0) (1/sinx-1/tanx) = lim_(x->0) (1/sinx-cosx/sinx)#

#lim_(x->0) (1/sinx-1/tanx) = lim_(x->0) (1-cosx)/sinx#

Now use the formulas for the double angle, considering #x = 2*x/2#:

#(1-cosx) = 1 - (cos^2(x/2) - sin^2(x/2)) = 1 - cos^2(x/2) + sin^2(x/2)) =2sin^2(x/2)#

#sinx = 2sin(x/2)cos(x/2)#

so that:

#lim_(x->0) (1/sinx-1/tanx) = lim_(x->0) (2sin^2(x/2))/(2sin(x/2)cos(x/2)) = lim_(x->0) tan(x/2) = 0#

Answer 2 · 2017-03-27T19:34:57

#lim_(xrarr0) (1/sinx-1/tanx) = lim_(xrarr0) sinx/(1+cosx) = 0#

#1/sinx-1/tanx = 1/sinx - cotx = 1/sinx-cosx/sinx = (1-cosx)/sinx#

# = ((1-cosx))/sinx * ((1+cosx))/((1+cosx))#

# = (1-cos^2x)/(sinx(1+cosx))#

# = sin^2x/(sinx(1+cosx))#

# = sinx/(1+cosx)#

So,

#lim_(xrarr0) (1/sinx-1/tanx) = lim_(xrarr0) sinx/(1+cosx)#

# = 0/(1+1) = 0#