Question

Find the value of ${lim}_{x\to 3}\frac{x^2-8x+15}{x^2-9x+18}$

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. $\frac{4}{3}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A

$\frac{2}{3}$

If substitute x=3 in the given expression we will get $\frac{0}{0}$. This is one of the indeterminate forms. First thing

we will try to do is factorize the given expression and cancel the common terms.

$\frac{x^2-8x+15}{x^2-9x+18}=\frac{(x-3)(x-5)}{(x-3)(x-6)}$

$=\frac{x-5}{x-6}$ if $x\ne 3$

$\underset{x\to 3}{lim}\frac{x^3-8x+15}{x^2-9x+18}=\underset{x\to 3}{lim}\frac{x-5}{x-6}$

$=\frac{-2}{-3}$

$=\frac{2}{3}$

We can cancel $x-3$ because $x=3\pm △x$ when we find the limit. Another way of solving this is by replacing

x with 3+h and letting h approach zero.

$\Rightarrow \underset{h\to 0}{lim}\frac{(h+3{)}^2-8(h+3)+15}{(h+3{)}^2-9(h+3)+18}$

$=\underset{h\to 0}{lim}\frac{h^2+6h+9-8h-24+15}{h^2+6h+9-9h-27+18}$

$=\underset{h\to 0}{lim}\frac{h^2-2h}{h^2-3h}=\underset{h\to 0}{lim}\frac{h-2}{h-3}$

$=\frac{2}{3}$
This is the right hand limit we calculated. To calculate the left hand limit we'll substitute x = 3 - h and on substituting that we'll find the same answer.
Since RHL and LHL are finite and equal. Limits exists and equal to 2 /3