show me the working of these maths questions?

Question

1) lim of (tan x -x)/x^3 as x approaches 0

2) lim of sqrt(4x^2+8x-5)-2x as x approaches infinity

Answer 1

1. The series expansion of tan x = x + x^3/3 + 2x^5/15 + (a bunch of terms with higher exponents of x).

Since you are taking the limit as x approaches 0, and dividing by x^3, terms with higher degree of x than 3 will not contribute to the result.

Subtracting x, because the numerator is "tan(x) - x", and ignoring terms of a higher degree than 3 leaves x^3/3. Dividing by x^3 yields 1/3 as the limit.

2. sqrt(4x^2 + 8x - 5) converges on sqrt((2x + 2)^2) as x approaches infinity.

Since the coefficient of x^2 is 4, the square root will have coefficient of x, sqrt(4)x = 2x, and be of the form "2x + z."

The constant term becomes irrelevant, so it's just a matter of figuring out what constant term would produce the "+8x"). You need to find z where (2x+z)^2 = (4x^2 + 8x + [some constant]).

Multiplying out, we get 8 = 2z + 2z, which reduces to z=2.

So, sqrt(4x^2+8x-5) converges to sqrt((2x+2)^2), or (2x+2), as x approaches infinity. Subtracting 2x leaves 2.

Answer 2

for the first one I'm going to assume it means tan (x) -x and not tan (x-x)

I would l'hopital rule it
(sec²(X)*1)/3x²
still gives you an indeterminate form so do it again
now we do some chain rule actions to get the derivative of
sec²(X).
Bring the exponent down and then multiply by the derivative of the inside
2sec²(x)tan(x)
then 3x² goes to 6x, which still doesn't help us
so 2sec²(x)tan(x)/6x
l'hopital it again
4sec²x tanx+sec²x(2sec²(x))
1/4cos²x *sinx/cosx
sinx/(cosx)^3+(1/cos²x)1/(2cos²x) --> 1/2(cosx)^4
sinx/(cosx)^3
multiply by 2cosx/2cosx giving you
(sinx 2cosx+1)/2(cosx)^4
and well that is really messy, but lets just leave like that for now and we will have that over 6 (because dy/dx 6x = 6)
((sinx 2cosx+1)/2(cosx)^4)/6
ok now lets just plug in 0 for x giving you
sin(x)->0 so the 2cosx doesn't matter
1/2(cos(0)^4
cos(0)->1
1^4=1
1*2=2
.5/6
so the limit as x-> zero is 1/12

I hope that works, I may have made a mistake somwhere in there.