Question

f(x) = (a/x) + ((bsinx)/(x)) + ((cx)/(pi)) + ((dx^2)/(pi^2))
what is lim x-> 0 for f(x)?
(i'll type the equation in the formula editor now to make it look clearer)

Answer 1

yes please

Answer 2

use \frac{top}{bottom} for fractions

Answer 3

\[\lim_{x \rightarrow 0} \frac{a}{x} + \frac{bsinx}{x} + \frac{cx}{\pi} + \frac{dx^2}{\pi^2}\]

Answer 4

thank you for the code i needed it! was looking for it everywhere

Answer 5

what is a?

Answer 6

probably a positive integer

Answer 7

where a, b, c, d is an unknown value

Answer 8

the answer can be in terms of a, b, c, d

Answer 9

the limit DNE if \[a\neq0\]

Answer 10

if a=0 the answer is b

Answer 11

I'm just taking the limit

Answer 12

yes i agree a has to be 0 for the limit to exist
and if a=0, then the limit is b
i agree zarkon
gj :)

Answer 13

oh i though you said L=0

Answer 14

may i ask why a = 0? sorry :(

Answer 15

cannot equal 0

Answer 16

can b = 0?

Answer 17

oops i meant \[\neq\]

Answer 18

i'm really confused.. so in order for the limit to exist a cannot be 0 and therefore b is 0?

Answer 19

\[\lim(b\cdot \frac{\sin(x)}{x})+\lim(\frac{a \pi^2+cx^2\pi+dx^3}{\pi^2x})\]
\[=b(1)+\lim \frac{a \pi^2+cx^2\pi+dx^3}{\pi^2x}\]
so for that second limit to exist we need it to be 0/0 when we plug in 0
but we get \[\frac{a \pi^2}{0}\]
s0 this implies a=0 in order for limit to exist
and if a=0 we have
\[=b+\lim \frac{cx^2 \pi+dx^3}{\pi^2 x}=b+\lim \frac{2xc \pi +3dx^2}{\pi^2}=b+\frac{0}{\pi^2}=b\]

Answer 20

thank you so much! the answer's perfect :) thank youuu!!