How do you evaluate #lim_(x->0) (sqrt(4+x)-2)/(3x)#?
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Given: #lim_(x to 0)(sqrt(4+x)-2)/(3x)#
Because the limit yields the indeterminant form #0/0#, one should use L'Hôpital's Rule. But, because the topic is "Determining Limits Algebraically", I shall assume that the student has not, yet, learned L'Hôpital's Rule and refrain from using it.
Multiply the expression by 1 in the form of #(sqrt(4+x)+2)/(sqrt(4+x)+2)#:
#lim_(x to 0)(sqrt(4+x)-2)/(3x)(sqrt(4+x)+2)/(sqrt(4+x)+2)#
The numerator becomes the difference of two squares:
#lim_(x to 0)((sqrt(4+x))^2-(2)^2)/((3x)(sqrt(4+x)+2))#
Expand the squares:
#lim_(x to 0)(4+x-4)/((3x)(sqrt(4+x)+2))#
Simplify the numerator:
#lim_(x to 0)x/((3x)(sqrt(4+x)+2))#
#x/x# becomes 1:
#lim_(x to 0)1/(3(sqrt(4+x)+2))#
Now, we may evaluate at #x = 0#:
#1/(3(sqrt(4)+2)) =1/12#
This limit is the same as the original expression:
#lim_(x to 0)(sqrt(4+x)-2)/(3x) = 1/12#
#lim_(x->0) (sqrt(4+x)-2)/(3x) = 1/12#
Apply L'Hospital's Rule :
#lim_(x->0) (sqrt(4+x)-2)/(3x)#
#= lim_(x->0) ((d((4+x)^(1/2)-2))/(dx))/((d(3x))/(dx))#
#=lim_(x->0) (1/2(4+x)^(-1/2))/(3)#
#=lim_(x->0) 1/(6sqrt(4+x))#
#=1/(6sqrt(4))#
#=1/12#
