L=Lim tan(x)^2/x x->0
Since both numerator and denominator evaluate to zero, we could apply l'Hôpital rule by taking derivatives.
d(tan^2(x))/dx=2tan(x).d(tan(x))/dx = 2tan(x)sec^2(x)
d(x)/dx = 1
=>
L=2tan(x)sec^2(x)/1 x->0
= (2(0)/1^2)/1
=0/1
=0
Another way using series,
We know that tan(x) = x+x^3/3+2x^5/15+.....
then tan^2(x), using binomial expansion gives
x^2+2*x^4/3+.... (we only need two terms)
and again apply l'Hôpital's rule, we have
L=d(x^2+2x^4/3+...)/d(x) = (2x+8x^3/3+...)/1
=0 as x->0
4.6 Is the answer I thinkkkkk
Answer:
6. (a-4)(a+9)
7. (3p+4)(p-2)
8. Will explain below.
9. (5b+4)(5b-4)
Step-by-step explanation:
HELLOOO IM BACK
6. -4*9 = -36 (the variable), 9-4 = 5 (a) , so a^2 +5a - 36 is fulfilled.
7. 4*-2 = -8, 4-2 = 2, so 3p^2 - 2p - 8 is fulfilled.
9. 25b^2 = (5b)^2
16 = 4^2
Using the difference of squares formula, a^2-b^2= (a+b)(a-b)
2:3
Given Racheal = 24 years
Thomson = 2/3×24
= 2×8
= 16 years
Must click thanks and mark brainliest
Answer:the answer is b
Step-by-step explanation:
