voltron21
  • voltron21
evaluate limx->0 (2.5+3.5x)^3 -15.625/ x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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ganeshie8
  • ganeshie8
expand the cube
voltron21
  • voltron21
and then what...
ganeshie8
  • ganeshie8
then u wil see that x cancels out top and bottom

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More answers

voltron21
  • voltron21
does it cancel all the x's in the numerator?
ganeshie8
  • ganeshie8
nopes, all we bother about is denominator x here, cuz denominator cant equal 0
ganeshie8
  • ganeshie8
pen down, things wil become clear...
voltron21
  • voltron21
im still lost
ganeshie8
  • ganeshie8
expand the cube first
voltron21
  • voltron21
ok i did that but i dont really see where ur going with this
ganeshie8
  • ganeshie8
wat do u get after expanding cube ?
voltron21
  • voltron21
(2.5+3.5x) three times
voltron21
  • voltron21
am i supposed to multiply that all together??
ganeshie8
  • ganeshie8
yes, expand it completely use this formula :- \(\large (a+b)^3=a^3+3a^2b+3ab^2+b^3 \)
ganeshie8
  • ganeshie8
\(\large \lim_{x->0} \frac{(2.5+3.5x)^3 -15.625}{x}\) \(\large \lim_{x->0} \frac{(2.5)^3 + 3(2.5)^2(3.5x) + 3(2.5)(3.5x)^2 + (3.5x)^3 -15.625}{x}\) \(\large \lim_{x->0} \frac{15.625 + 3(2.5)^2(3.5x) + 3(2.5)(3.5x)^2 + (3.5x)^3 -15.625}{x}\)
ganeshie8
  • ganeshie8
\(\large \lim_{x->0} \frac{\cancel{15.625} + 3(2.5)^2(3.5x) + 3(2.5)(3.5x)^2 + (3.5x)^3 \cancel{-15.625}}{x}\)
ganeshie8
  • ganeshie8
\(\large \lim_{x->0} \frac{3(2.5)^2(3.5x) + 3(2.5)(3.5x)^2 + (3.5x)^3 }{x}\)
ganeshie8
  • ganeshie8
notice that, you can factor out x from entire numerator, and cancel the denominator
voltron21
  • voltron21
ok thanks for all the help i got 65.625... wanna help me with 2 more problems?
ganeshie8
  • ganeshie8
\(\large \lim_{x->0} \frac{x [3(2.5)^2(3.5) + 3(2.5)(3.5^2x) + (3.5^3x^2)] }{x}\) \(\large \lim_{x->0} \frac{\cancel{x} [3(2.5)^2(3.5) + 3(2.5)(3.5^2x) + (3.5^3x^2)] }{\cancel{x}}\)
ganeshie8
  • ganeshie8
65.625 is \(\large \color{Red}{\checkmark}\) good work !
ganeshie8
  • ganeshie8
yea sure :)
voltron21
  • voltron21
youre lovely!
voltron21
  • voltron21
lim x-0 X^2/sin^2(5x)
ganeshie8
  • ganeshie8
aww thnks :) that keeps me going for next two problems lol
voltron21
  • voltron21
lol so i changed sin^2 to 1-cos
voltron21
  • voltron21
but idk what to do now because that doesnt equal 0
ganeshie8
  • ganeshie8
we use this :- \(\large \lim_{x->0} \frac{\sin(x)}{x} = 1\)
ganeshie8
  • ganeshie8
convert given limit to this form
voltron21
  • voltron21
so the answer would be 1?
ganeshie8
  • ganeshie8
nopes, in the given limit, we dont have same x top and bottom. so you need to rearrange a bit
voltron21
  • voltron21
so do i have to flip it? like multiply by the inverse?
ganeshie8
  • ganeshie8
Yes
ganeshie8
  • ganeshie8
\(\large \lim_{x->0} \frac{x^2}{\sin^2 5x}\) \(\large \lim_{x->0} (\frac{x}{\sin 5x})^2\) \(\large \lim_{x->0} (\frac{5x}{5\sin 5x})^2\) \(\large \frac{1}{25} \lim_{x->0} (\frac{5x}{\sin 5x})^2\)
ganeshie8
  • ganeshie8
now flip it, and take the limit inside the exponent
ganeshie8
  • ganeshie8
\(\large \frac{1}{25} \lim_{x->0} (\frac{1}{\frac{\sin 5x}{5x}})^2\)
ganeshie8
  • ganeshie8
as x->0, 5x->0
ganeshie8
  • ganeshie8
\(\large \frac{1}{25} (\frac{1}{ \lim_{5x->0} \frac{\sin 5x}{5x}})^2\)
ganeshie8
  • ganeshie8
\(\large \frac{1}{25} (\frac{1}{1})^2\)
voltron21
  • voltron21
ok so those are essentially go to 0 so the answer is 1/25 got it!
ganeshie8
  • ganeshie8
\(\large \frac{1}{25} \)
voltron21
  • voltron21
\[f(x)=\sqrt{3x^2+4} / 5x+3\] find hte horizontal asymptote
voltron21
  • voltron21
\[f(x)= \frac{ \sqrt{3x^2+4} }{ 5x+3 }\]

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