Question 219650
If we take the direct route, we get:



{{{lim(x->7,(sqrt(x+2)-3)/(x-7))}}} Start with the given limit.



{{{(sqrt(7+2)-3)/(7-7)}}} Plug in {{{x=7}}} (let's just pretend that we can evaluate this limit the intuitive way).



{{{(sqrt(9)-3)/(7-7)}}} Add



{{{(3-3)/(7-7)}}} Take the square root of 9 to get 3.



{{{(0)/(0)}}} Subtract



Since the above is indeterminate, we have to take another approach...


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{{{lim(x->7,(sqrt(x+2)-3)/(x-7))}}} Start with the given limit.


Because there's a radical in the numerator, we can try to rationalize the numerator in hopes that something will cancel.



{{{lim(x->7,((sqrt(x+2)-3)(sqrt(x+2)+3))/((x-7)(sqrt(x+2)+3)))}}} Multiply both the numerator and denominator by the conjugate of {{{sqrt(x+2)-3}}}



{{{lim(x->7,((x+2)-9)/((x-7)(sqrt(x+2)+3)))}}} FOIL the numerator (use the difference of squares formula).



{{{lim(x->7,(x-7)/((x-7)(sqrt(x+2)+3)))}}} Combine like terms.



{{{lim(x->7,cross((x-7))/(cross((x-7))(sqrt(x+2)+3)))}}} Cancel out the common factors.



{{{lim(x->7,1/(sqrt(x+2)+3))}}} Simplify. Now let's see if we can evaluate this limit using the naive approach (ie use {{{lim(x->a,f(x))=f(a)}}}).



{{{1/(sqrt(7+2)+3)}}} Plug in {{{x=7}}}



{{{1/(sqrt(9)+3)}}} Add



{{{1/(3+3)}}} Take the square root of 9 to get 3.



{{{1/6}}} Add



So {{{lim(x->7,(sqrt(x+2)-3)/(x-7))=1/6}}}



An alternative (and better way in my opinion) is to use L' Hospital's Rule. If you haven't learned that yet, then ignore that last suggestion (note: the rule only works if you're dealing with interderminate forms).