Question

At the limit approaches 0 the limit equals (tanx-x)/x^3

Answer 1

@freckles

Answer 2

k we can use l'hospital again if you want

Answer 3

since we have 0/0 when plugin in 0

Answer 4

Yes I like that rule

Answer 5

did you know how to diff (tan(x)-x) and x^3?

Answer 6

tanx is secx I think?!?! than -x=-1 So (secx-1)/3x^2 I think?

Answer 7

sec^2(x)=(tan(x))'

Answer 8

\[\lim_{x \rightarrow 0}\frac{\sec^2(x)-1}{3x^2}\]

Answer 9

Okay is tan the only one that is ^2?

Answer 10

what do you mean?

Answer 11

It's been so long I feel like since we did the basic differentiation

Answer 12

like tan is sec^2x is that the only one that has the squared in it?

Answer 13

the derivative of cot(x) is -csc^2(x)

Answer 14

Okay, ill write that down

Answer 15

derivative of sin(x) is cos(x)
derivative of cos(x) is -sin(x)
derivative of tan(x) is sec^2(x)
derivative of csc(x) is -csc(x) cot(x)
derivative of sec(x) is sec(x) tan(x)
derivative of cot(x) is -csc^2(x)

Answer 16

Omg thank you! :)

Answer 17

but anyways we have 0/0 again when pluggin in 0 so you can do another round of l'hospital

Answer 18

the only thing I see you might have trouble with is diff sec^2(x)
you will need chain rule for this

Answer 19

sec^2(x) is the same as writing (sec(x))^2

Answer 20

\[\frac{d}{dx}(\sec(x))^2=2(\sec(x))^{2-1} \cdot (\sec(x))'\]

Answer 21

okay so there will be 2(sec(x))*sec(x)tan(x)

Answer 22

that is right
or 2 sec^2(x) tan(x)

Answer 23

\[\lim_{x \rightarrow 0} \frac{2 \sec^2(x) \tan(x)}{6x}\]

Answer 24

we could do another round of l'hospital or rearrange this one

Answer 25

your call

Answer 26

lets just do another round

Answer 27

than it will help me with the basics again

Answer 28

\[\lim_{x \rightarrow 0} \frac{2 \sec^2(x) \frac{\sin(x)}{\cos(x)}}{6x} \\ =\lim_{x \rightarrow 0} \frac{2 \sec^2(x)}{\cos(x)} \frac{\sin(x)}{6x} \\ =\frac{2}{6}\lim_{x \rightarrow 0} \frac{ \sec^2(x)}{\cos(x)} \frac{\sin(x)}{x}\]

Answer 29

see if you can manage the limit from here

Answer 30

wait it you did DS wouldn't your answer just be 2/6 or 1/3?

Answer 31

*if

Answer 32

2/6 reduces to 1/3 yes
you have just after my last line the following:
\[\frac{2}{6} \cdot \frac{1^2}{1} \cdot 1 =\frac{2}{6}=\frac{1}{3}\]

Answer 33

as x approaches 0:

sin(x)/x goes to 1
sec^2(x) goes to 1 since sec^2(0) is 1
cos(x) goes to 1 since cos(1) is 1
so sec^2(x)/cos(x)=1

Answer 34

oops mistype

Answer 35

cos(0) is 1*

Answer 36

Note to jot down lol

Answer 37

Thank you so much :)