Maths questions. Need help!?

1. [cos(sin x) - 1 ] * sin(^2)x / (x^2 * sin(^2)x)

n we know lim x-->0 [cos(sin x) - 1] / sin(^2)x = - 1/2

n sin^x / x =1

so ans would be... = -(1/2) * 1^2

= - 1/2

2. this would be of the form 1^infinity...

so ans = e ^[lim x->0 (x+1)*(1/x)]

= e^ [ 1/1] ( by differentiating numerator n denominator... L'hospital rule )

= e --------------->answer

3. Since 2y = x^2 implies y = (1/2)x^2, a point on the curve in terms of x is (x, (1/2)x^2).

The distance between (x, (1/2)x^2) and (0,5) is:

S = √{[(1/2)x^2 - 5]^2 + (x - 0)^2} = √[(1/4)x^4 - 4x^2+ 25].

Taking derivatives yields:

dS/dx = (x^3 - 8 x)/{2√[(1/4)x^4 -4x^2+ 25]}.

Setting dS/dx = 0:

(x^3 - 8x)/{2√[(1/4)x^4 + 25]} = 0

==> x^3 - 8x= 0

==> x = 0 , +2√2 , - 2√2

==> y = 0 , 4 , 4

check 2nd derivative---

3x^2 - 8 =0

for minimum distance ...2nd derivative should b +ve..

so for x= 0 its -ve...

n for x= +2√2 and -2√2 its +ve...

Therefore, the point on the curve closest to (0,5) is (+2√2,4) or (-2√2,4).

1.

as x -> 0 sin x ∼ x

lim(x-->0) [cos(sin x) -1]/x^2 = lim(x-->0) [cos(x) -1]/x^2

hopital rule

lim(-sin x)/2x

hopital rule again

lim (-cos x)/2 --> -1/2 (as x --> 0)

2.

lim (x+1)^(1/x) = (substitute x = 1/t)

x->0

lim (1 + 1/t)^t = e

t->inf

f(0) = e

3.

the curve can be parametrized as

(t, (1/2)t^2)

distance from point (0,5) is

f(t) = √(t^2 + ((1/2)t^2 - 5)^2) = (1/2)√(t^4 - 16t^2 + 100)

distance is minimum when

g(t) = t^4 - 16t^2 + 100

is minimum where g'(t) = 0 and g''(t) > 0

g '(t) = 4t^3 - 32t = 0

t=0, t = ±√8

g ''(t) = (t^6 - 24t^4 + 300t^2 - 800)/(t^4 - 16t^2 + 100)^(3/2)

g ''(0) = -4/5 --> relative maximum

g ''(±√8) = 8/3 > 0 --> relative minima at

(-√8,4) and (√8,4)