How do you compute the limit of #cot(4x)/csc(3x)# as #x->0#?

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Answer 1 · 2017-03-06T23:59:35

#= 3/4#

# lim_(x to 0) cot(4x)/csc(3x)#

#=lim_(x to 0) ( cos(4x) sin(3x))/(sin (4x) #

#=lim_(x to 0) cos(4x) ( 3x(sin(3x))/(3x))/(4x(sin (4x))/(4x)) #

#=lim_(x to 0) cos(4x) ( 3(sin(3x))/(3x))/(4(sin (4x))/(4x)) #

The limit of a product/quotient is the product/quotient of the limits:

#=lim_(x to 0) cos(4x) cdot ( 3lim_(x to 0) (sin(3x))/(3x))/(4lim_(x to 0) (sin (4x))/(4x)) #

Now:

#lim_(z to 0) (sin z)/z = 1#

And:

#=lim_(z to 0) cos(z) = 1#

So we have:

#=1 cdot ( 3 cdot 1)/(4 cdot 1) = 3/4 #