Question

evaluate limx->0 (2.5+3.5x)^3 -15.625/ x

Answer 1

expand the cube

Answer 2

and then what...

Answer 3

then u wil see that x cancels out top and bottom

Answer 4

does it cancel all the x's in the numerator?

Answer 5

nopes, all we bother about is denominator x here, cuz denominator cant equal 0

Answer 6

pen down, things wil become clear...

Answer 7

im still lost

Answer 8

expand the cube first

Answer 9

ok i did that but i dont really see where ur going with this

Answer 10

wat do u get after expanding cube ?

Answer 11

(2.5+3.5x) three times

Answer 12

am i supposed to multiply that all together??

Answer 13

yes, expand it completely
use this formula :-
\(\large (a+b)^3=a^3+3a^2b+3ab^2+b^3 \)

Answer 14

\(\large \lim_{x->0} \frac{(2.5+3.5x)^3 -15.625}{x}\)

\(\large \lim_{x->0} \frac{(2.5)^3 + 3(2.5)^2(3.5x) + 3(2.5)(3.5x)^2 + (3.5x)^3 -15.625}{x}\)

\(\large \lim_{x->0} \frac{15.625 + 3(2.5)^2(3.5x) + 3(2.5)(3.5x)^2 + (3.5x)^3 -15.625}{x}\)

Answer 15

\(\large \lim_{x->0} \frac{\cancel{15.625} + 3(2.5)^2(3.5x) + 3(2.5)(3.5x)^2 + (3.5x)^3 \cancel{-15.625}}{x}\)

Answer 16

\(\large \lim_{x->0} \frac{3(2.5)^2(3.5x) + 3(2.5)(3.5x)^2 + (3.5x)^3 }{x}\)

Answer 17

notice that, you can factor out x from entire numerator, and cancel the denominator

Answer 18

ok thanks for all the help i got 65.625... wanna help me with 2 more problems?

Answer 19

\(\large \lim_{x->0} \frac{x [3(2.5)^2(3.5) + 3(2.5)(3.5^2x) + (3.5^3x^2)] }{x}\)

\(\large \lim_{x->0} \frac{\cancel{x} [3(2.5)^2(3.5) + 3(2.5)(3.5^2x) + (3.5^3x^2)] }{\cancel{x}}\)

Answer 20

65.625 is \(\large \color{Red}{\checkmark}\)

good work !

Answer 21

yea sure :)

Answer 22

youre lovely!

Answer 23

lim x-0 X^2/sin^2(5x)

Answer 24

aww thnks :) that keeps me going for next two problems lol

Answer 25

lol so i changed sin^2 to 1-cos

Answer 26

but idk what to do now because that doesnt equal 0

Answer 27

we use this :-

\(\large \lim_{x->0} \frac{\sin(x)}{x} = 1\)

Answer 28

convert given limit to this form

Answer 29

so the answer would be 1?

Answer 30

nopes, in the given limit, we dont have same x top and bottom. so you need to rearrange a bit

Answer 31

so do i have to flip it? like multiply by the inverse?

Answer 32

Yes

Answer 33

\(\large \lim_{x->0} \frac{x^2}{\sin^2 5x}\)

\(\large \lim_{x->0} (\frac{x}{\sin 5x})^2\)

\(\large \lim_{x->0} (\frac{5x}{5\sin 5x})^2\)

\(\large \frac{1}{25} \lim_{x->0} (\frac{5x}{\sin 5x})^2\)

Answer 34

now flip it, and take the limit inside the exponent

Answer 35

\(\large \frac{1}{25} \lim_{x->0} (\frac{1}{\frac{\sin 5x}{5x}})^2\)

Answer 36

as x->0, 5x->0

Answer 37

\(\large \frac{1}{25} (\frac{1}{ \lim_{5x->0} \frac{\sin 5x}{5x}})^2\)

Answer 38

\(\large \frac{1}{25} (\frac{1}{1})^2\)

Answer 39

ok so those are essentially go to 0 so the answer is 1/25 got it!

Answer 40

\(\large \frac{1}{25} \)

Answer 41

\[f(x)=\sqrt{3x^2+4} / 5x+3\] find hte horizontal asymptote

Answer 42

\[f(x)= \frac{ \sqrt{3x^2+4} }{ 5x+3 }\]