Question #59763

3 Answers
Andrew I.
Jan 12, 2018

#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

turksvids
Jan 12, 2018

#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)#.

Jim H
Jan 12, 2018

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#

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