Question

lim/deltax approaches 0 = 1/(cos(x + deltax)) - 1/(cosx)

Answer 1

SOLVED but connection failed.

Answer 2

\[\lim_{\delta x \rightarrow 0}\left( \frac{ 1 }{\cos \left( x+\delta x \right) }-\frac{ 1 }{\cos x } \right)\]

Answer 3

That is the problem exactly.

Answer 4

Well...not exactly. The stuff in the parentheses is all over delta x

Answer 5

\[=\lim_{\delta x \rightarrow 0}\frac{ \cos x-\cos \left( x+\delta x \right) }{\cos \left( x+\delta x \right) \cos x }\]

Answer 6

\[\cos C-\cos D=2\sin \frac{ C+D }{ 2 }\sin \frac{ D-C }{2 }\]
Now you can solve if any problem,show me your work i will guide you.

Answer 7

HOw does this change if the whole thing is over delta x?

Answer 8

do not solve denominator.
If you have to divide by delta x.
Make it frac{ sin frac{ delta x }{2 } }{frac{ delta x }{2 }*2 } and take limits.

Answer 9

I'm calling \(\Delta x=h\) to save some time:
\[\lim_{h\to0}\frac{\dfrac{1}{\cos{(x+h)}}-\dfrac{1}{\cos x}}{h}\]
\[\lim_{h\to0}\frac{\dfrac{\cos x}{\cos x\cos{(x+h)}}-\dfrac{\cos(x+h)}{\cos x\cos(x+h)}}{h}\]
\[\lim_{h\to0}\frac{1}{h}\left(\frac{\cos x-\cos(x+h)}{\cos x\cos(x+h)}\right)\]
From the angle sum identity for cosine, you have
\[\cos(x+y)=\cos x\cos y-\sin x\sin y\]
\[\lim_{h\to0}\frac{1}{h}\left(\frac{\cos x-\cos x\cos h+\sin x\sin h}{\cos x(\cos x\cos h-\sin x\sin h)}\right)\]
\[\lim_{h\to0}\frac{1}{h}\left(\frac{\cos x-\cos x\cos h}{\cos x(\cos x\cos h-\sin x\sin h)}\right)+\lim_{h\to0}\frac{1}{h}\left(\frac{\sin x\sin h}{\cos x(\cos x\cos h-\sin x\sin h)}\right)\]
\[\lim_{h\to0}\frac{1-\cos h} {h}\left(\frac{\cos x}{\cos x(\cos x\cos h-\sin x\sin h)}\right)\\~~~~~~~~~~~~+\lim_{h\to0}\frac{\sin h}{h}\left(\frac{\sin x}{\cos x(\cos x\cos h-\sin x\sin h)}\right)\]
\[\color{red}{\lim_{h\to0}\frac{1-\cos h} {h}}\left(\frac{1}{\cos x\cos h-\sin x\sin h}\right)+\color{blue}{\lim_{h\to0}\frac{\sin h}{h}}\left(\frac{\tan x}{\cos x\cos h-\sin x\sin h}\right)\]
You should already be familiar with the red and blue limits. First term disappears, and the second term leaves you with
\[\lim_{h\to0}\frac{\tan x}{\cos x\cos h-\sin x\sin h}\]

Answer 10

\[=\lim_{\delta x \rightarrow 0}\frac{ 2\sin \frac{ x+x+\delta x }{2 }\sin \frac{ x+\delta x-x }{ 2 } }{\delta xcos \left( x+\delta x \right)\cos x }\]

Answer 11

\[=\lim_{\delta x \rightarrow 0}\frac{ 2\sin \left( x+\frac{ \delta x }{ 2 } \right)\sin \frac{ \delta x }{ 2 } }{2* \frac{ \delta x }{ 2 } \cos \left( x+\delta x \right) \cos x }\]

Answer 12

\[=\frac{ \sin \left( x+0 \right)*1 }{\cos \left( x+0 \right)\cos x }=\sec x \tan x\]

Answer 13

\[\because \lim_{\delta x \rightarrow 0} \frac{ \frac{\sin \delta x }{ 2 } }{ \frac{ \delta x }{ 2 } } =1\]

Answer 14

in the beginning i have not divided by delta x, because i didn't know what you wanted to do.