10 points to who can help me with these calculus limits problems?

Question

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lim (3(a^2) + sin(4a))/(a)

a -> 0

lime (sin^2(a/2))/(a^2)

a -> 0

thank you soo much!!!

Answer 1

  lim (3a² + sin 4a)/a

a→0

= lim (3a²/a) + (sin 4a)/a

  a→0

= lim { (3a²/a) + (sin 4a)/a }

  a→0

= lim {(3a) + 4[ (sin 4a)/4a ]}

  a→0

= 0 + 4

= 4

Answer 2

Develop sinx according    sin x=x-x^3/3 (x->0)

1)sin 4a= 4a-64a^3/3

then lim(3a^2+sin4a)/a=3a+4+64a^2/3=4

2) sin(a/2)= a/2 -a^3/6      sin^(~)= a^2/4- a^4/6

lim(sin^2(a/2))/a^2=1/4-a^2/6=1/4

Answer 3

Using L'Hopital's Rule,

Differentiate the numerator and the denominator

lim a ->0 (6a + 4 cos(4a)) / 1

As a->0, 6a->0

As a->0, cos(4a) -> 1

The limit is 4

Use L'Hopital's Rule.

lim a->0 2 sin(a/2) cos(a/2) (1/2)  /  2a

lim a->0 sin(a/2) cos(a/2) / 2a

differentiate again

lim a->0 [-sin(a/2) sin(a/2)(1/2) + cos(a/2)cos(a/2) (1/2)] /2

As a->0, sin(a/2)->0, cos(a/2) ->1

as a->0, the numerator approaches 1/2 , so the limit is

1/2 divided by 2 = 4

Answer 4

first sin(x)/x when x-> to 0 is 1

so you multiply  the sin by 4/4 and get lim (3(a^2) + [sin(4a))/(4a)]*4

lim (3(a^2) + 4 =4  (0^2=0)

Answer 5

We know that lim x->0 (sin x)/x = 1:

lim x->0 (sin x)/x

= lim x->0 (cos x)/1 (using l'Hopital's rule)

= cos(0)/1 = 1/1 = 1

Using this, we can solve the two limits as follows:

lim (3a² + sin(4a))/a

a>0

= lim 3a²/a + (sin(4a))/a

= lim 3a + lim 4(sin(4a))/4a

= lim 3a + 4 lim (sin(4a))/4a

= 0 + 4x1

= 4

lim (sin²(a/2))/a²

a>0

= lim (sin(a/2))/a x (sin(a/2))/a

= lim (sin(a/2))/(2(a/2)) x (sin(a/2))/(2(a/2))

= lim (1/2)(sin(a/2))/(a/2) x (1/2)(sin(a/2))/(a/2)

= (1/2) lim (sin(a/2))/(a/2) x (1/2) lim (sin(a/2))/(a/2)

= (1/2)(1) x (1/2)(1)

= 1/4

NOTE: How do we know that  lim a->0 (sin(4a))/4a = 1 ?

Let x = 4a, then as a->0, 4a->0 therefore x->0

Replacing 4a by x, and a->0 by x->0, we get

lim a->0 (sin(4a))/4a = lim x->0 (sin x)/x  = 1 (as found above)

Similarly, by letting x = a/2, then

lim a->0 (sin(a/2))/(a/2) = lim x->0 (sin x)/x = 1

Answer 6

using L´Hôpital

1) =lim 6a+4cosa =4

2) = (sin(a/2)*cos(a/2))/2a ===>1/4

Answer 7

a) lim a-->0   [6a+4cos(4a)] = 4

b) lima-->0 [sin[a/2]cos[a/2]]/2a

   =lima-->0 sina/4a = 1/4

Answer 8

lim (3(a^2) + sin(4a))//a = lim (3(a^2)/a + sin(4a))/a = 0 + 4 = 4

lim (sin^2(a/2))/(a^2) = lim (sin(a/2)/a)^2 = 0.5^2=0.25

a -> 0