What is the limit of #(sqrt(x+5) - 3)/(x - 4)# as x approaches 4?
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2 Answers
Let's multiply both numerator and denominator of this expression by #sqrt(x+5)+3# to get rid of undefined #0/0# value.
Assuming #x!=4# in our expression,
#(sqrt(x+5)-3) / (x-4) = #
#= [(sqrt(x+5)-3) * (sqrt(x+5)+3)] / [(x-4)*(sqrt(x+5)+3)] =#
#= (x+5-9) / [(x-4)*(sqrt(x+5)+3)] =#
#= (x-4)/ [(x-4)*(sqrt(x+5)+3)] =#
Eazydraw 9 5 2 X 4 10
#= 1/ (sqrt(x+5)+3)#
As #x->4#, this expression tends to #1/(sqrt(4+5)+3)=1/6#.
Therefore,
#lim_(x->4)(sqrt(x+5)-3) / (x-4) = 1/6#
Explanation:
#lim_{x->4}(sqrt{x+5}-3)=0#
#lim_{x->4}(x-4)=0#
Jacuzzi rooms downtown detroit. Using the L'Hospital Rule,
Let's multiply both numerator and denominator of this expression by #sqrt(x+5)+3# to get rid of undefined #0/0# value.
Assuming #x!=4# in our expression,
#(sqrt(x+5)-3) / (x-4) = #
#= [(sqrt(x+5)-3) * (sqrt(x+5)+3)] / [(x-4)*(sqrt(x+5)+3)] =#
#= (x+5-9) / [(x-4)*(sqrt(x+5)+3)] =#
#= (x-4)/ [(x-4)*(sqrt(x+5)+3)] =#
Eazydraw 9 5 2 X 4 10
#= 1/ (sqrt(x+5)+3)#
As #x->4#, this expression tends to #1/(sqrt(4+5)+3)=1/6#.
Therefore,
#lim_(x->4)(sqrt(x+5)-3) / (x-4) = 1/6#
Explanation:
#lim_{x->4}(sqrt{x+5}-3)=0#
#lim_{x->4}(x-4)=0#
Jacuzzi rooms downtown detroit. Using the L'Hospital Rule,
#lim_{x->4}frac{sqrt{x+5}-3}{x-4}#
Eazydraw 9 5 2 X 4 3
#=lim_{x->4}frac{frac{d}{dx}(sqrt{x+5}-3)}{frac{d}{dx}(x-4)}#
#=lim_{x->4}frac{(frac{1}{2sqrt{x+5}})}{1}#
Coushatta casino events calendar. #=frac{1}{2sqrt{4+5}}#
Eazydraw 9 5 2 X 4 9
#=1/6#
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Eazydraw 9 5 2 X 4
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