Question

Question #59763

Answer 1

`tan(x)sec(x)`

Explanation:

To start:

`lim_(Deltax->0)(sec(x+Deltax)-sec(x))/(Deltax)`

`=lim_(Deltax->0)(1/cos(x+Deltax)-1/cos(x))/(Deltax)`

`=lim_(Deltax->0)((cos(x)-cos(x+Deltax))/(cos(x)cos(x+Deltax)))/(Deltax)`

`=lim_(Deltax->0)(cos(x)-cos(x+Deltax))/(cos(x)cos(x+Deltax)Deltax)`

Now use the trig identity:

`cos(A)-cos(B)=-2sin((A+B)/2)sin((A-B)/2)`

So we now have:

`=lim_(Deltax->0)(-2sin((x+x+Deltax)/2)sin((x - x-Deltax)/2))/(cos(x)cos(x+Deltax)Deltax)`

`=lim_(Deltax->0)(-2sin(x+(Deltax)/2)sin(-(Deltax)/2))/(cos(x)cos(x+Deltax)Deltax)`

Take the negative at the front and cancel with the negative inside the second sin function and multiply the top and the bottom by `1/2` to take the factor of `2` at the top and place on the bottom like so:

`=lim_(Deltax->0)(sin(x+(Deltax)/2)sin((Deltax)/2))/(cos(x)cos(x+Deltax)(Deltax)/2)`

`=lim_(Deltax->0)(sin(x+(Deltax)/2))/(cos(x)cos(x+Deltax))*lim_(Deltax->0)sin((Deltax)/2)/((Deltax)/2)`

`lim_(Deltax->0)sin((Deltax)/2)/((Deltax)/2)=1`

So we are left with:

`=lim_(Deltax->0)(sin(x+(Deltax)/2))/(cos(x)cos(x+Deltax))`

Evaluating the limit by direct substitution:

`=sin(x)/(cos(x)cos(x))=sin(x)/cos(x)*1/cos(x)=tan(x)sec(x)`

That is the derivative of `sec(x)` so that is exactly what we expected.

For a proof that:

`lim_(Deltax->0)sin((Deltax)/2)/((Deltax)/2)=1`

follow this link:
https://socratic.org/questions/how-do-you-use-the-squeeze-theorem-to-find-lim-sin-x-x-as-x-approaches-zero

Answer 2

`sec(x)tan(x)`

Explanation:

To evaluate `lim_(Deltax to 0)((sec(x+Deltax)-sec(x))/(Deltax))` notice that is of the form `lim_(Deltax to 0)((f(x+Deltax)-f(x))/(Deltax))`, which is the limit definition of the derivative.

So,

`lim_(Deltax to 0)((sec(x+Deltax)-sec(x))/(Deltax))=f'(x)`,

where `f(x)=sec(x)`.

For `f(x)` it's known that `f'(x)=sec(x)tan(x)` so:

`lim_(Deltax to 0)((sec(x+Deltax)-sec(x))/(Deltax))=sec(x)tan(x)`.

Answer 3

Very much like Andrews I.'s answer. The details are different.

Explanation:

We'll need:
`sect=1/cost` and

`cos(x+Deltax)=cosx cos Deltax - sinx sin Deltax`

and the limits:

`lim_(Deltaxrarr0)(sin Deltax)/(Deltax)=1` and `lim_(Deltaxrarr0)(1-cos Deltax)/(Deltax)=0`.

Solution

`lim_(Deltaxrarr0)(sec(x+Deltax)-secx)/(Deltax) = lim_(Deltaxrarr0)(1/(cos(x+Deltax))-1/cos(x))/(Deltax)`

`= lim_(Deltaxrarr0)((cosx-cos(x+Deltax))/(Deltaxcos(x)cos(x+Deltax)))`

Expand `cos(x+Deltax)` in the numerator only. (Its would be OK to expand it in the denominator, but it is not necessary.)

`= lim_(Deltaxrarr0)(cosx-(cosxcos Deltax-sinxsin Deltax))/(Deltaxcos(x)cos(x+Deltax))`

Regroup to use the fundamental trigonometric limits.

`= lim_(Deltaxrarr0)[(cosx-cosxcos Deltax)/(Deltaxcosxcos(x+Deltax))+(sinxsin Deltax)/(Deltaxcos(x)cos(x+Deltax))]`

`= lim_(Deltaxrarr0)[cosx/(cosxcos(x+Deltax))((1-cos Deltax)/(Deltax))+(sinx)/(cos(x)cos(x+Deltax))((sin Deltax)/(Deltax))]`

Evaluate the limit using the fundamental trig limits and continuity of cosine.

`= cosx/(cosxcos(x))(0)+(sinx)/(cos(x)cos(x))(1)`

`=sinx/cos^2x=1/cosxsinx/cosx=secxtanx`