How do you find the limit of #(x-4)/(sqrt(x-3)-sqrt(5-x))# as x approaches 4?

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Answer 1 · 2015-04-13T18:25:58

The answer is: #1#.

This limit shows up in the #0/0# indecision form, so we have to rationalise.

#lim_(xrarr4)(x-4)/(sqrt(x-3)-sqrt(5-x))=#

#=lim_(xrarr4)[(x-4)/(sqrt(x-3)-sqrt(5-x))*(sqrt(x-3)+sqrt(5-x))/(sqrt(x-3)+sqrt(5-x))]=#

#=lim_(xrarr4)[(x-4)(sqrt(x-3)+sqrt(5-x))]/(x-3-5+x)=#

#=lim_(xrarr4)[(x-4)(sqrt(x-3)+sqrt(5-x))]/(2x-8)=#

#=lim_(xrarr4)[(x-4)(sqrt(x-3)+sqrt(5-x))]/[2(x-4)]=#

#=lim_(xrarr4)[(sqrt(x-3)+sqrt(5-x))]/2=(sqrt(4-3)+sqrt(5-4))/2=1#.

Answer 2 · 2015-04-13T18:27:36

Rationalize the denominator by multiplying the fraction by #1# in the form:

#(sqrt(x-3)+sqrt(5-x))/(sqrt(x-3)+sqrt(5-x))#

This works because #(a-b)(a+b) = a^2-b^2#,

so #(sqrta-sqrtb)(sqrta+sqrtb) = a-b#. It looks like this:

#((x-4))/((sqrt(x-3)-sqrt(5-x))) ((sqrt(x-3)+sqrt(5-x)))/((sqrt(x-3)+sqrt(5-x)))#

# =((x-4) (sqrt(x-3)+sqrt(5-x)))/((x-3)-(5-x)#

# =((x-4) (sqrt(x-3)+sqrt(5-x)))/(2x-8)#

# =((x-4) (sqrt(x-3)+sqrt(5-x)))/(2(x-4))#

# =(sqrt(x-3)+sqrt(5-x))/2#

So, we have:

#lim_(x rarr 4) (x-4)/(sqrt(x-3)-sqrt(5-x))=lim_(xrarr4) (sqrt(x-3)+sqrt(5-x))/2 #

# = 2/2=1#