Math / Calculus

Question $\lim _{x \rightarrow(-2)} \frac{2-|x|}{x+2}$

Studdy Solution

STEP 1

1. The function involves an absolute value, which may affect the limit from different directions.
2. We need to consider the behavior of the function as $x$ approaches $-2$ from both the left and the right.
3. The limit exists only if the left-hand limit and the right-hand limit are equal.

STEP 2

1. Analyze the behavior of the absolute value function around $x = -2$ .
2. Calculate the left-hand limit as $x \to -2^-$ .
3. Calculate the right-hand limit as $x \to -2^+$ .
4. Compare the left-hand and right-hand limits to determine if the overall limit exists.

STEP 3

First, understand how the absolute value function $|x|$ behaves around $x = -2$ . - For $x < 0$ , $|x| = -x$ . - For $x \geq 0$ , $|x| = x$ .

STEP 4

Calculate the left-hand limit as $x \to -2^-$ (approaching $-2$ from the left):
For $x < -2$ , $|x| = -x$ . Substitute this into the function:
$\lim_{x \to -2^-} \frac{2 - |x|}{x + 2} = \lim_{x \to -2^-} \frac{2 - (-x)}{x + 2} = \lim_{x \to -2^-} \frac{2 + x}{x + 2}$
As $x \to -2^-$ , both the numerator and denominator approach 0. This indicates a need for further simplification or analysis, such as factoring or recognizing a form of indeterminate limit.

STEP 5

Simplify the expression $\frac{2 + x}{x + 2}$ :
Notice that $2 + x = x + 2$ , so:
$\lim_{x \to -2^-} \frac{2 + x}{x + 2} = \lim_{x \to -2^-} \frac{x + 2}{x + 2} = \lim_{x \to -2^-} 1 = 1$

STEP 6

Calculate the right-hand limit as $x \to -2^+$ (approaching $-2$ from the right):
For $x \geq -2$ , $|x| = -x$ . Substitute this into the function:
$\lim_{x \to -2^+} \frac{2 - |x|}{x + 2} = \lim_{x \to -2^+} \frac{2 - (-x)}{x + 2} = \lim_{x \to -2^+} \frac{2 + x}{x + 2}$
As $x \to -2^+$ , the expression simplifies similarly:
$\lim_{x \to -2^+} \frac{2 + x}{x + 2} = \lim_{x \to -2^+} \frac{x + 2}{x + 2} = \lim_{x \to -2^+} 1 = 1$

STEP 7

Compare the left-hand and right-hand limits:
Since both the left-hand limit and the right-hand limit as $x \to -2$ are equal to 1, the overall limit exists and is:
$\lim_{x \to -2} \frac{2 - |x|}{x + 2} = 1$
The value of the limit is:
$\boxed{1}$

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