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Math  /  Calculus

Questiononcepts Araph of a function h mathematics, the graph of a function is the set of ordered pairs, where in the common ce find the end behavior in limit notation f(x)=x−1x2−4f(x)=\frac{x-1}{x^{2}-4}

Studdy Solution

STEP 1

What is this asking? We need to describe what happens to the function f(x)f(x) as xx gets really big (approaches infinity) and really small (approaches negative infinity). Watch out! Don't just plug in a large number for *x*!
We need to use limit notation and understand the behavior of the function.

STEP 2

1. Analyze the function
2. Find the limit as *x* approaches infinity
3. Find the limit as *x* approaches negative infinity

STEP 3

Alright, let's **break down** this function!
We've got f(x)=x−1x2−4f(x) = \frac{x - 1}{x^2 - 4}.
Notice that the **highest power** of xx in the numerator is xx (which is x1x^1), and the **highest power** of xx in the denominator is x2x^2.

STEP 4

This tells us something super important!
As xx becomes absolutely massive (either positively or negatively), the x2x^2 term in the denominator will **grow much faster** than the xx term in the numerator.
Imagine plugging in x=1000x = 1000.
The numerator is close to 1000, but the denominator is close to 1,000,000!
The denominator is way bigger, so the overall fraction will be very small.

STEP 5

To **formalize** this, we'll use limit notation.
We want to find lim⁡x→∞f(x)=lim⁡x→∞x−1x2−4\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{x - 1}{x^2 - 4}.

STEP 6

A handy trick here is to **divide** both the numerator and the denominator by the highest power of xx in the denominator, which is x2x^2.
This is like multiplying by 1x21x2\frac{\frac{1}{x^2}}{\frac{1}{x^2}}, which is just multiplying by 1, so it doesn't change the value of the expression!

STEP 7

So we get: lim⁡x→∞xx2−1x2x2x2−4x2=lim⁡x→∞1x−1x21−4x2\lim_{x \to \infty} \frac{\frac{x}{x^2} - \frac{1}{x^2}}{\frac{x^2}{x^2} - \frac{4}{x^2}} = \lim_{x \to \infty} \frac{\frac{1}{x} - \frac{1}{x^2}}{1 - \frac{4}{x^2}}

STEP 8

Now, as xx approaches infinity, 1x\frac{1}{x} approaches **zero**, 1x2\frac{1}{x^2} approaches **zero**, and 4x2\frac{4}{x^2} approaches **zero**.
So, we have: lim⁡x→∞1x−1x21−4x2=0−01−0=01=0\lim_{x \to \infty} \frac{\frac{1}{x} - \frac{1}{x^2}}{1 - \frac{4}{x^2}} = \frac{0 - 0}{1 - 0} = \frac{0}{1} = 0

STEP 9

We do the **same thing** for negative infinity!
We want to find lim⁡x→−∞f(x)=lim⁡x→−∞x−1x2−4\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{x - 1}{x^2 - 4}.

STEP 10

We **divide** both the numerator and denominator by x2x^2 again: lim⁡x→−∞xx2−1x2x2x2−4x2=lim⁡x→−∞1x−1x21−4x2\lim_{x \to -\infty} \frac{\frac{x}{x^2} - \frac{1}{x^2}}{\frac{x^2}{x^2} - \frac{4}{x^2}} = \lim_{x \to -\infty} \frac{\frac{1}{x} - \frac{1}{x^2}}{1 - \frac{4}{x^2}}

STEP 11

As xx approaches negative infinity, 1x\frac{1}{x} still approaches **zero**, 1x2\frac{1}{x^2} approaches **zero**, and 4x2\frac{4}{x^2} approaches **zero**.
So, we have: lim⁡x→−∞1x−1x21−4x2=0−01−0=01=0\lim_{x \to -\infty} \frac{\frac{1}{x} - \frac{1}{x^2}}{1 - \frac{4}{x^2}} = \frac{0 - 0}{1 - 0} = \frac{0}{1} = 0

STEP 12

As xx approaches infinity, f(x)f(x) approaches **0**.
As xx approaches negative infinity, f(x)f(x) also approaches **0**.
In limit notation: lim⁡x→∞f(x)=0\lim_{x \to \infty} f(x) = 0 and lim⁡x→−∞f(x)=0\lim_{x \to -\infty} f(x) = 0.

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