Question

Is the answer 24?
Q)If a,b,c,d, are constants that
lim x->0 (ax^2+sinbx+sincx+sindx)/(3x^2+5x^2+7x^6)=8
find the value of a+b+c+d !

Answer 1

0/0, apply L'H rule. but i'm stuck on the 2nd part, where it turns out (0+b+c+d)/0 ,,,i can't apply L''H rule!

Answer 2

did you write the denominator right?

Answer 3

can i apply the limit to x. so plug in 0, then from the step
(0+b+c+d)/0 , set it equal to 8. then cross multiply, and get b+c+d=0?
it make sense if thats true. bcuz in the end, i get 24 for a. so the whole answer is 24

Answer 4

holy shoot no! it's (3x^2+5x^4+7x^6) sry!

Answer 5

any clues :l?

Answer 6

i'm still finding the trick to the problem

Answer 7

i kept on applying LH rule. but it can't b applied after the first one....

Answer 8

i have some. still figuring out. i ran out of scratch. i'll share it, that you may find some solution\[\huge \lim_{x \rightarrow 0}\frac{ \frac{ ax^3 }{ 3 }-\frac{ \cos bx }{ b } -\frac{ \cos cx }{ c }-\frac{ \cos dx }{ d }}{ x^3+x^5+x^7 }=\left[ \frac{ 0 }{ 0 } \right]\]

Answer 9

ur going backwards! should b 2ax+bsinbx+....etc

Answer 10

3x^2 + 5x^4 + 7x^6 is the derivative of x^3 + x^5 + x^7. so maybe the problem was a result of an application of L'HR.

Answer 11

rnt u suppose to take L'HR again n again til u get it.

Answer 12

let's apply the L'HR once.
\[\large =\lim_{x \rightarrow 0}\frac{ 2ax+b \cos bx + c \cos cx + d \cos dx }{ 6x+20x^3+42x^5 }\]

Answer 13

yes im there. then plug in 0 for x right?

Answer 14

(0+b+c+d)/0 :(

Answer 15

sirm~~ plz check your inbox. can't physically stay up xD 1hr 50min sleep left. grrrr

Answer 16

ok. i'm still solving.

Answer 17

thank you so much : ) !

Answer 18

\[\large = \lim_{x \rightarrow 0}\frac{ 2ax+b+c+d+b \cos bx - b+ c \cos cx -c + d \cos dx -d }{ 2x(3+10x^2+21x^4) }\]

Answer 19

regrouping terms, \[=\lim_{x \rightarrow 0}\frac{ 2ax+b+c+d }{ 2x(3+10x^2+21x^4) }+\frac{ b }{ 6 }\lim_{x \rightarrow 0}\frac{ \cos bx - 1 }{ x }+\frac{ b }{ 6 }\lim_{x \rightarrow 0}\frac{ \cos cx - 1 }{ x }+\frac{ d }{ 6 }\lim_{x \rightarrow 0}\frac{ \cos dx - 1 }{ x }\]

Answer 20

\[\large =\lim_{x \rightarrow 0}\frac{ 2ax+b+c+d }{ 2x(3+10x^2+21x^4) }=8\]because the the limit of the original expression is 8.

Answer 21

IF\[\large b+c+d \neq 0\]then \[\large \lim_{x \rightarrow 0}\frac{ 2ax+b+c+d }{ 2x(3+10x^2+21x^4) }=\infty\]contradicting the assumption that the limit is 8.
THUS, a + b + c = 0 or \[\large \lim_{x \rightarrow 0}\frac{ 2ax }{ 2x(3+10x^2+21x^2) }=\frac{ a }{ 3 }=8\]

Answer 22

oops, that's b + c + d = 0, NOT a + b + c = 0

Answer 23

THEREFORE\[\large a=24 \text{ and }\space a+b+c+d=24\]