How do you solve #lim_x->pi/4 (tanx-cotx)/(x-pi/4)# ?
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L'Hospital Rule yields:
#lim_(x->pi/4)(tanx-cotx)/(x-pi/4)=4#
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 :)
Here is a Solution without use of the L'Hospital's Rule.
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#.
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#
