Math / Calculus

Question[3] $\lim _{x \rightarrow 2^{-}} \frac{x+2}{x^{3}+8}$

Studdy Solution

STEP 1

What is this asking? We need to find the limit of a fraction as $x$ approaches $\text{2}$ from the left. Watch out! Plugging in $x = 2$ directly gives us a zero in the denominator, so we need to be careful!

STEP 2

1. Factor the Denominator
2. Analyze the Limit

STEP 3

Alright, let's factor that denominator!
We have $x^3 + 8$ , which is a sum of cubes.
Remember the sum of cubes formula?
It's $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ .
Here, $a = x$ and $b = 2$ , so $x^3 + 8 = (x+2)(x^2 - 2x + 4)$ .

STEP 4

So, our expression becomes $\frac{x+2}{(x+2)(x^2 - 2x + 4)}.$

STEP 5

Now, since we're taking the limit as $x$ approaches $\text{2}$ (and not at $x = 2$ ), $x$ is close to $\text{2}$ , but not equal to it.
This means $x + 2$ is not zero, so we can divide the numerator and denominator by $x + 2$ .
This gives us $\frac{1}{x^2 - 2x + 4}.$ Remember, we're looking at values of $x$ that are close to $\text{2}$ , but not equal to $\text{2}$ , so dividing by $x + 2$ is perfectly safe!

STEP 6

Now we can analyze the limit as $x$ approaches $\text{2}$ from the left.
Since we've simplified our expression, we can just plug in $x = 2$ into our new expression: $\frac{1}{2^2 - 2 \cdot 2 + 4}.$

STEP 7

Let's evaluate the denominator: $2^2 - 2 \cdot 2 + 4 = 4 - 4 + 4 = 4$ .

STEP 8

So, the limit is $\frac{1}{4}.$ Woohoo! We did it!

STEP 9

The limit of the expression as $x$ approaches $\text{2}$ from the left is $\frac{1}{4}$ .

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