Evaluate the following limit algebraically: 

 Apr 21, 2020

a direct substitution gives you 0/0, which means more work can be done

let's simplify the numerator by subtracting the two fractions and see what we get

\(lim_{x\rightarrow 3}\frac{\frac{1}{6}-\frac{1}{9-x}}{x-3}\\ =lim_{x\rightarrow 3}\frac{\frac{9-x}{6(9-x)}-\frac{6}{6(9-x)}}{x-3}\\ =lim_{x\rightarrow 3}\frac{\frac{9-x-6}{6(9-x)}}{x-3}\\ =lim_{x\rightarrow 3}\frac{\frac{-x+3}{6(9-x)}}{x-3}\)

all i did was subtract the two fractions by giving them both common denominators. in this case, i multiplied the numerator and denominator of 1/6 by 9-x and the numerator and denominator of 1/9-x by 6. This gives the two fractions a common denominator of 6(9-x).

Let's keep going
\(=lim_{x\rightarrow 3}\frac{\frac{-(x-3)}{6(9-x)}}{x-3}\\ =lim_{x\rightarrow 3}\frac{-(x-3)}{x-3}*\frac{1}{6(9-x)}\\ =lim_{x\rightarrow 3}{-1}*\frac{1}{6(9-x)}=-1*\frac{1}{6(9-3)}=-\frac{1}{36}\)

i factored out the negative from -x+3 and moved away 6(9-x) to make it more noticeable that x-3 can be canceled out. this allows for direct substitution of x=3 which gives us -1/36

 Apr 21, 2020

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