Question

Question #086cf

Answer 1

`= 2`

Explanation:

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`

Answer 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`