How do you solve lim_x->pi/4 (tanx-cotx)/(x-pi/4) ?

3 Answers

L'Hospital Rule yields:

lim_(x->pi/4)(tanx-cotx)/(x-pi/4)=4

Explanation:

by L'Hospital rule, which states that: when you get an indeterminate form after assuming that lim_(x->c)f(x)=f(c), then lim_(x->c)(g(x))/(h(x))=(g'(c))/(h'(c))

=> lim_(x-> pi/4) ((secx)^2+ (cscx)^2)/1

placing values of x=pi/4

= ( (sqrt(2)) ^2 + (sqrt(2))^2 ) / 1 = 4/1 = 4

you can find the value of the limit as now it is not in the indertiminate form

hope you find it helpful :)

Aug 14, 2018

Here is a Solution without use of the L'Hospital's Rule.

Explanation:

Let, x=pi/4+y.

:." As "x to pi/4, y to 0.

Now, tanx-cotx=sinx/cosx-cosx/sinx,

=(sin^2x-cos^2x)/(sinxcosx),

={-2(cos^2x-sin^2x)}/{2sinxcosx},

=(-2cos2x)/(sin2x),

=-2cot2x.

:." The Reqd. Lim."=lim_(x to pi/4)(-2cot2x)/(x-pi/4),

=lim_(y to 0){-2cot(2(pi/4+y))}/y,

=lim_(y to 0){-2cot(pi/2+2y)}/y,

=lim_(y to 0){-2(-tan2y)}/y,

=lim_(y to 0){2*(tan2y)/(2y)*2}.

Knowing that, lim_(theta to 0)tantheta/theta=1,

"The Lim."=4*1=4.

Aug 14, 2018

Please see below.

Explanation:

Solution without L'Hospital Rule :

Let,

L=lim_(x to pi/4) (tanx-cotx)/(x-pi/4)

Here,

tanx-cotx=sinx/cosx-cosx/sinx=(sin^2x-cos^2x)/(sinxcosx)

:.tanx-cotx=-(cos^2x-sin^2x)/(2sinxcosx)xx2

:.tanx-cotx=-2(cos2x)/(sin2x)=-2cot2x

So,

L=lim_(x to pi/4)(-2cot2x)/((x-pi/4))=-2lim_(x to pi/4)(cot2x)/((x-pi/4))

Let , x-pi/4=theta=>x=pi/4+theta=>2x=pi/2+2theta

and x->pi/4=>theta->0

:.L=-2lim_(thetato0)(cot(pi/2+2theta))/theta=-2lim_(theta to0)(-tan2theta)/theta

:.L=2lim_(theta to0)(tan2theta)/(2theta )xx2=2(1)xx2=4