What is the limit for the following?

a) #lim_(x->0)(sin6x)/(sin7x)#

I would try some Pre-Calculus identities here. Forget the limit for now, because it's just #0/0#, and focus on the argument.

#sin6x = sin(3x+3x) = sin3xcos3x+cos3xsin3x#
#= 2sin3xcos3x#

#sin7x = sin(4x+3x) = sin4xcos3x+cos4xsin3x#

So we have:

#(2sin3xcos3x)/(sin4xcos3x+cos4xsin3x)#

#= (2sin3x)/(sin4x+((cos4x)/(cos3x))sin3x)#

#= (2)/((sin4x)/(sin3x)+(cos4x)/(cos3x))#

By factoring out and canceling #cos3x# and then #sin3x#, it seems a little more manageable now.

Instead of having to work with #sinnx# AND #cosnx# (which would be really complicated), we only have to work with simplifying those #sinnx# because the #cosnx# are together now. Now, let's focus on the #sinnx#.

#(sin4x)/(sin3x) = (sin(2x+2x))/(sin(2x+x)) = (2sin2xcos2x)/(sin2xcosx+cos2xsinx)#

More identities to know:

#sin2x = 2sinxcosx# (you've seen this)
#cos2x = cos^2x - sin^2x = 1-2sin^2x = 2cos^2x-1#

Plug them in, work with it a bit, and realize that there are #sinx# factors in common. Sweet!

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

#= (2[4(sinx)cos^3x-2(sinx)cosx])/[[(sinx)(2cos^2x)+(sinx)(2cos^2x-1)]]#

Now we can factor out #sinx# and simplify some more!

#= cancel((sinx)/(sinx))((8cos^3x-4cosx))/((2cos^2x)+(2cos^2x-1))#

#= ((8cos^3x-4cosx))/(4cos^2x-1)#

Returning to the original expression:

#lim_(x->0) (2)/(((8cos^3x-4cosx))/(4cos^2x-1)+(cos4x)/(cos3x))#

Now we can plug in 0 to get:

#= lim_(x->0) (2)/(((8(1^3)-4(1)))/(4(1^2)-1)+(1/1)#

#= lim_(x->0) (2)/((4)/(3)+1) = 2/(7/3) = 6/7#

b) #lim_(x->0)e^(-2x)sqrtx#

There is nothing you need to do here. Just plug in #0# to get:

#e^(0)sqrt(0) = 1*0 = 0#

c) #lim_(x->5)(x^2-5x)/(x-5)#

This one is simple too. Factor and cancel:

#= lim_(x->5)[xcancel{(x-5)}]/cancel{(x-5)}#

#=> lim_(x->5) x = 5#