Question

!"#
Δ! → 0
Still need help!
lim = sec (x + Δx) − sec (x)
Δx--0 Δx

Answer 1

you will have to retype this in the comments using the equation button I dont even know exactly what is going on

Answer 2

are you saying\[\lim_{\Delta x \rightarrow 0}=\frac{ \sec(x+\Delta x)-\sec(x) }{ \Delta x }\]

Answer 3

yes!

Answer 4

ok so first multiply by the conjugate of \[\sec(x+\Delta x)+\sec(x)\] what do you get

Answer 5

sec^2(x + delta x) - sec^2 x

Answer 6

or wait sorry still in the mindset of this other problem I have to take another look at this

Answer 7

change your sec to 1/cos and simplify that

Answer 8

I changed them to 1/cos and got a common denominator to combine them
\[cosx - \cos(x + Deltax) \over \cos(x + deltax)cosx\]

Answer 9

I didn't think I could cancel the cos because they are being subtracted in the numerator. Did I miss something?

Answer 10

I dont understand what you did when you changed to 1/cos because you would have\[\frac{ 1 }{ \cos(x+\Delta x)-\cos(x) }*\frac{ 1 }{ \Delta x }\]

Answer 11

and sorry you were right about the cos

Answer 12

@agent0smith we need your help

Answer 13

yeah I'm getting stumped myself

Answer 14

\[\frac{ 1 }{ cosx + \delta x } - \frac{ 1 }{ cosx }\times \frac{ 1 }{ Deltax }\]
Then I got a common denominator between the two fractions.

Answer 15

oh wait I see now

Answer 16

since it is as dleta x approaches zero you can plug in zero for delta x now and simplify

Answer 17

it should simplify the fraction more for you

Answer 18

Will that give me zero in the denominator?

Answer 19

plug it in\[\frac{ \cos(x)-\cos(x-0) }{ \cos(x+0)\cos(x) }\]

Answer 20

so you now have \[\frac{ \cos(x)-\cos(x) }{ \cos ^{2}(x) }\]

Answer 21

I still have the 1/delta x also from the original problem since it was all over delta x to begin with. If I replace delta x with zero, does that give me zero in the denominator?

Answer 22

show me where it would be in the equation I'm really sorry but I am cooking dinner at the same time so you will have to do most of the heavy lifting

Answer 23

Bless your heart for doing calculus AND dinner! The original problem was completely over delta x. We inverted and multiplied leaving a delta x in the denominator. I was hoping that somehow the delta x's would cancel out, but I have tried every identity and trick I know and cannot get rid of them.

Answer 24

I'm guessing so far you've changed the secants into 1/cosine, and then used the angle subtraction formulas... i'll work it out myself though

Answer 25

thanks agent0smith we need your help we've kind of been all over the place

Answer 26

I used the angle addition of Cos(A +B) = CosACosB -SinASinB --then I kind of got stuck.

Answer 27

After some simplifying it looks like \[\Large \frac{ \cos x - \cos x \cos \Delta x+ \sin x \sin \Delta x }{ \cos x( \cos x \cos \Delta x - \sin x \sin \Delta x) } * \frac{ 1 }{\Delta x }\]

The result should eventually be secxtanx, since we're finding the derivative of secx...

Answer 28

I got this far as well, but I must be missing something in the simplification of this. How did you get to secxtanx? Did you split it into two expressions and work them separately?

Answer 29

I didn't get to secxtanx. I said that the derivative of secx is secxtanx.

Answer 30

OH..sorry, I misunderstood. What is the limit of this problem as delta x approaches zero?

Answer 31

I haven't figured out how to show the work yet, but the original problem is basically finding the derivative of secx (you might be too early into calculus/precalc to know this yet), so your eventual solution will be secxtanx.

Answer 32

Thank you so much for your help. I haven't had calculus for a lot of years and I am trying to look up what I do not remember! I am sure it will come back to me with some practice. If you think of a way to show how the problem simplifies--that would be great. Thanks again to both of you recon14193 and Agent0smith

Answer 33

I tried using wolfram alpha to simplify it, no help :/

Answer 34

Hmm, if i divide the numerator by cosx i'll get a tanx up top \[\Large \frac{ \cos x - \cos x \cos \Delta x+ \sin x \sin \Delta x }{ \cos x( \cos x \cos \Delta x - \sin x \sin \Delta x) } * \frac{ 1 }{\Delta x }\]

Answer 35

After dividing the numerator and denominator by cos x \[\Large \frac{1 - \cos \Delta x+ \tan x\sin \Delta x }{ \cos x \cos \Delta x - \sin x \sin \Delta x } * \frac{ 1 }{\Delta x }\]

Answer 36

Still don't see the Delta x disappearing anytime soon but i'll try to figure it out....

Answer 37

Maybe this will help http://www.youtube.com/watch?v=ClGVHlpx9_A

Answer 38

now lets rewrite \[\lim_{\Delta x \rightarrow 0}\left( \frac{ 1-\cos \Delta x +\tan x \sin \Delta x}{ \cos x \cos \Delta x-\sin x \sin \Delta x }\frac{ 1 }{ \Delta x } \right)=\frac{ \lim_{\Delta x \rightarrow 0}\frac{ 1-\cos \Delta x+\tan x \sin \Delta x }{ \Delta x } }{ \lim_{\Delta x \rightarrow 0}\left( \cos x \cos \Delta x -\sin x \sin \Delta x\right) }\]
now you can figure out the denominator by substituting 0 Delta x, cos(0)=1 and sin(0)=0, so denominator becomes 1/cos(x) or sec(x). now lets figure out top equation:\[\lim_{\Delta x \rightarrow 0}\frac{ 1-\cos \Delta x+\tan x \sin \Delta x }{ \Delta x }=\lim_{\Delta x \rightarrow 0}\frac{ 1-\cos \Delta x }{ \Delta x }+\tan x \lim_{\Delta x \rightarrow 0}\frac{ \sin \Delta x }{ \Delta x}=\tan x\] because \[\lim_{\Delta x \rightarrow 0}\frac{ 1-\cos \Delta x }{ \Delta x }=0\] and\[\lim_{\Delta x \rightarrow 0}\frac{ \sin \Delta x }{ \Delta x }=1\] so your limit becomes \[\sec x \tan x\]

Answer 39

if you need help prooving those 2 limits i gave you answers, I'm here to help

Answer 40

Is there anyway to prove lim (1-cosx)/x = 0 w/o l'hopitals rule? That's what i was trying to avoid, any use of that (i feel like otherwise you can just use it right from the start). Good solution regardless.

Answer 41

yes there is\[\lim_{\Delta x \rightarrow 0}\frac{ 1- \cos \Delta x }{ \Delta x }=\lim_{\Delta x \rightarrow 0}\left( \frac{ 1-\cos \Delta x }{ \Delta x }\frac{ 1+\cos \Delta x }{ 1+\cos \Delta x } \right)=\lim_{\Delta x \rightarrow 0}\frac{ 1-\cos^2 \Delta x }{ \Delta x \left( 1+\cos \Delta x \right) }=\lim_{\Delta x \rightarrow 0}\frac{ \sin^2 \Delta x }{ \Delta x \left( 1+\cos \Delta x \right) }=\lim_{\Delta x \rightarrow 0}\frac{ \sin \Delta x }{ \Delta x }\lim_{\Delta x \rightarrow 0}\frac{ \sin \Delta x }{ 1+\cos \Delta x }\]

Answer 42

and you prove sin(Delta x)/(Delta x)=1 by squeezing theorem

Answer 43

l'hopital's rool is easiest way out

Answer 44

or just differentiating 1/cosx by chain rool