Question

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

Answer 1

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 :)

Answer 2

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`.

Answer 3

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`