Question

lim(x→∞)⁡(1-2x^2)/((4x+3)^2)

Thank you Solomon, you are awesome. I have one more I need help with. Thank you again

Answer 1

\(\large\color{slate}{\displaystyle\lim_{v \rightarrow ~\infty}\frac{1-2x^2}{(4x+3)^2}}\) this ?

Answer 2

I mean, \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty}\frac{1-2x^2}{(4x+3)^2}}\) . Like this ?

Answer 3

I can't help you, if you don't even let me know if my interpretation is correct or not.

Answer 4

yes you are correct with the second entry, sorry had to take dog out

Answer 5

Expand the denominator by multiplying it out

Answer 6

that's about all I know how to do lol, just started this today so I am not to good at it yet

Answer 7

Okay, did you expand it?

Answer 8

16x^2+9

Answer 9

That isn't correct

Answer 10

you forgot the middle term, @ingallina

Answer 11

16x^2+24+9

Answer 12

Ok, I thought of...

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty}\frac{1-2x^2}{(4x+3)^2}}\)


\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty}\frac{-(2x^2-1)}{(4x+3)^2}}\)


\(\large\color{slate}{-\left[ \displaystyle\lim_{x \rightarrow ~\infty}\frac{(2x^2-1)}{(4x+3)^2}\right]}\)

then L'H'S twice

Answer 13

what does lhs twice mean? and I do not understand why you have same problem three different ways? I just need to solve it showing steps to help me understand

Answer 14

I mean to differentiate top and bottom of the limit twice

Answer 15

Do you understand how I got, \(\large\color{slate}{-\left[ \displaystyle\lim_{x \rightarrow ~\infty}\frac{(2x^2-1)}{(4x+3)^2}\right] }\)

Answer 16

??

Answer 17

ok

Answer 18

can you find the derivative of the top and of the bottom?

Answer 19

im so lost

Answer 20

I would expand on the bottom first.

Answer 21

how do you expand the \(\large\color{slate}{(4x+3)^2 }\) ?

Answer 22

16x^2+24+9

Answer 23

yes

Answer 24

24x you mean

Answer 25

re-writing it, \(\large\color{slate}{-\left[ \displaystyle\lim_{x \rightarrow ~\infty}~\frac{(2x^2-1)}{16x^2+24x+9}\right] }\)

Answer 26

find the derivative of top and bottom of the limit

Answer 27

Do you know how to find the derivative of each of the terms in the denominator and the numerator?

Answer 28

common number? is bottom 8(4x+3)????????

Answer 29

you are trying to find the derivative?

Answer 30

kay, lets start from here, what is the derivative of 2x^2 ?

Answer 31

(Use the power rule)

Answer 32

4x sorry it takes so long to reply I have to reconnect everytime

Answer 33

yes, I understand, correct, what is the derivative of 1?

Answer 34

1?

Answer 35

Line y=3 will always have a slope of ?

Answer 36

and any line of y=C, will have a slope of ?

Answer 37

yes

Answer 38

So, what is derivative of -1, and of 9 ?

Answer 39

0

Answer 40

yes,

Answer 41

i got to go in 10 minutes

Answer 42

on no!!! can you help me finish this please, can u list steps so I can learn form it..please

Answer 43

hello

Answer 44

@ingallina , Solomon was using L\hopital's rule, a simple way of solving limits, and since your function is in fractions, that's the best way to use l'hoptal's rule. If the limit is \(\large \frac{0}{0}\) and \(\large \frac{oo}{oo}\), these are the indeterminate forms. If you come up with an indeterminate answer for the limit, you then take the derivative of the top and bottom, if the result is still in indeterminate form, then take the second derivative, till you no longer have any indeterminate forms.

L'Hopital's rule states:
If the limit is \(\large \frac{0}{0}\) and \(\large \frac{oo}{oo}\)
then \(\large\lim_{x \rightarrow a}\frac{f(x)}{g(x)}\frac{\large\bf Use~LHOPITAL'S~rule}{=}\lim_{x \rightarrow a}\frac{f'(x)}{g'(x)}\)

Answer 45

Divide numerator and denominator by highest power term. In this case it is \(x^2\).