Question

Lim x^2 - 2x - 15/x^2-9

Answer 1

Answer:

5/3 if x → 0, 1 if x → ∞

Step-by-step explanation:

I'm going to answer this for two potential scenarios: the limit as x approaches 0, and the limit as x approaches infinity. First, let's factor and simplify our expression a little. $x^2-2x-15$ factors into $(x-5)(x+3)$, and $x^2-9$ is the difference of squares $x^2-3^2$, so we can factor it into $(x+3)(x-3)$. Our expression now becomes

$\dfrac{(x-5)(x+3)}{(x-3)(x+3)}=\dfrac{x-5}{x-3}$

If we take the limit of this as x → 0:

$\lim_{x\to0}\frac{x-5}{x-3}= \frac{-5}{-3}=\frac{5}{3}$

And the limit as x → ∞:

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

An explanation for that second limit: As x gets larger and larger, the 5 and 3 being subtracted in the numerator and denominator become less and less significant, and the fraction gets closer and closer to the value x/x, which is 1.