How to find a horizontal asymptote #(x^2 - 5x + 6)/( x - 3)#?

Question

Chloe Alexander · Accepted Answer

There is no horizontal asymptote

There is no horizontal asymptote as degree of numerator #2# is greater than that of denominator #1# by one. In such case, there is a possibility of a slant asymptote, but before concluding that let us factorize #(x^2−5x+6)# as follows: #(x^2−5x+6)=x^2-3x-2x+6=x(x-3)-2(x-3)=(x-2)(x-3)# Hence #(x^2−5x+6)/(x−3)# can be simplified as follows: #((x-2)(x-3))/(x-3)# or #(x-2)# Hence #(x^2−5x+6)/(x−3)# is the equation of just the line #y=(x-2)# (I do not think tis can be considered as slanting asymptote). But as #x-3# appears in denominator the domain of#x# in #y# does not include #3#.

Sadie Alexander · Answer

undefined To find the horizontal asymptote of the function $\frac{x^2 - 5x + 6}{x - 3}$, we examine the behavior of the function as $x$ approaches positive or negative infinity.

As $x$ approaches positive or negative infinity, the term $x - 3$ dominates the expression since the degree of the numerator and the denominator are the same. Therefore, we can divide each term in the numerator by $x$ to see what happens as $x$ approaches infinity:

$$ \lim_{x \to \infty} \frac{x^2 - 5x + 6}{x - 3} $$

$$ = \lim_{x \to \infty} \frac{x^2/x - 5x/x + 6/x}{x/x - 3/x} $$

$$ = \lim_{x \to \infty} \frac{x - 5 + \frac{6}{x}}{1 - \frac{3}{x}} $$

As $x$ approaches infinity, $\frac{6}{x}$ and $\frac{3}{x}$ approach 0, leaving us with:

$$ = \lim_{x \to \infty} \frac{x - 5 + 0}{1 - 0} $$

$$ = \lim_{x \to \infty} (x - 5) $$

Since the limit of $x - 5$ as $x$ approaches infinity is infinity, there is no horizontal asymptote. Instead, the function grows without bound as $x$ approaches positive infinity. Similarly, as $x$ approaches negative infinity, the function also grows without bound. Therefore, there is no horizontal asymptote for the given function.