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Solve the following (limit; L'Hospital's Rule)?

#\lim_(x\rightarrow0^+)(\tan(2x))^x#

What I've tried
#y=(\tan(2x))^x#
#\rArr\ln(y)=\lim_(x\rightarrow0^+)\ln((\tan(2x))^x)...#
got stuck afterwards.

Answer 1

Ezekiel Alexander

Of course, please provide the specific limit you would like to solve using L'Hôpital's Rule.

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Answer 2

William Abdalla

#1#

#(tan(2x))^x=(sin(2x))^x/(cos(2x))^x#

#sin(2x)= 2x-(2x)^3/(3!)+(2x)^5/(5!)+ cdots#

is an alternate series and for #-pi/2 le x le pi/2# verifies

#2x-(2x)^3/(3!) le sin(2x) le 2x#

then

#lim_(x->0)(2x-(2x)^3/(3!))^x le lim_(x->0)(sin(2x))^x le lim_(x->0)( 2x)^x#

now using the binomial expansion

#x^x = (1+(x-1))^x = 1 + x(x-1)+x(x-1)(x-1)^2/(2!)+x(x-1)(x-2)(x-1)^3/(3!)+cdots = 1+xp(x)# where #p(x)# represents a suitable polynomial. So

#lim_(x->0)x^x=lim_(x->0) 1+xp(x)=1#

Of course

#lim_(x->0)(2x)^x=(lim_(y->0)y^y)^(1/2)= 1^(1/2)=1#

Finally

#lim_(x->0)(tan(2x))^x=(lim_(x->0)(sin(2x))^x)/(lim_(x->0)(cos(2x))^x)=1/1^0= 1#

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Answer 3

Emilia Aguilar

Let y = (tan 2x)^x, then

#lim x to 0_+- ln y#

#= lim x to 0_+- x ln tan 2x#

# = lim x to 0_+- (ln tan 2x)/(1/x)#

#=lim x to 0_+- (( ln tan 2x )')/((1/x)')#

#=lim x to 0_+- -2x^2sec^2 (2x)/tan (2x) #

#=lim x to 0_+-((2x)/(sin 2x))((x))/((cos2x)#

#=(1)(0)/1=0#. so,

#y to e^0 = 1#.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Solve the following (limit; L'Hospital's Rule)?

#\lim_(x\rightarrow0^+)(\tan(2x))^x# What I've tried #y=(\tan(2x))^x# #\rArr\ln(y)=\lim_(x\rightarrow0^+)\ln((\tan(2x))^x)...# got stuck afterwards.

#\lim_(x\rightarrow0^+)(\tan(2x))^x#

#\lim_(x\rightarrow0^+)(\tan(2x))^x#

What I've tried
#y=(\tan(2x))^x#
#\rArr\ln(y)=\lim_(x\rightarrow0^+)\ln((\tan(2x))^x)...#
got stuck afterwards.