Question #086cf

I think you would be well advised to ask one at a time.

First one:

#lim_(x to 0) (tan x - x )/(x - sin x)#

This could be done using L'Hôpital's rule as it's #0/0# indeterminate, but I'll use a series expansion.

The series we need are:

#sin x approx x - x^3/(3!) + x^5/(5!) - x^7/(7!) + ... #

....and:

#tan x approx x + 1/3 x^3 + 2/15 x^5 + 17/315 x^7 + ...#

I'll put in a load of terms to make the point but typically only the first one or two terms are that relevant.

#implies lim_(x to 0) (1/3x^3 + 2/15 x^5 + 17/315 x^7 + ... )/( x^3/(3!) - x^5/(5!) + x^7/(7!) + ... )#

As we are looking at #lim_(x to 0)#, clearly we can ignore #mathcal O (x^5)# to get:

#implies lim_(x to 0) (x^3/(3) )/( x^3/(3!) ) = (3!)/3 = 2#

The fifth one:

#lim_(xrarr1)(1-x)tan((pix)/2)=lim_(xrarr1)((1-x)sin((pix)/2))/cos((pix)/2)#

This is in the indeterminate form #0/0#, so we can take the derivative of the numerator over the numerator of the denominator. We will use product and chain rules:

#=lim_(xrarr1)(-sin((pix)/2)+(1-x)(pi/2cos((pix)/2)))/(-pi/2sin((pix)/2))#

We can now plug in #x=1#:

#=(-sin(pi/2))/(-pi/2sin(pi/2))=color(blue)(2/pi#