Math / Calculus

QuestionFind the limit, if it exists. (If an answer does not exist, enter DNE.) $\lim _{x \rightarrow-\infty} \frac{x-4}{x^{2}+1}$

Studdy Solution

STEP 1

1. The expression involves a limit as $x$ approaches $-\infty$ .
2. The function to evaluate the limit is $\frac{x-4}{x^2+1}$ .
3. To find the limit, we need to analyze the behavior of both the numerator and the denominator as $x$ approaches $-\infty$ .

STEP 2

1. Analyze the behavior of the numerator $x-4$ as $x$ approaches $-\infty$ .
2. Analyze the behavior of the denominator $x^2 + 1$ as $x$ approaches $-\infty$ .
3. Simplify the expression $\frac{x-4}{x^2+1}$ by dividing both the numerator and the denominator by $x^2$ .
4. Evaluate the limit of the simplified expression as $x$ approaches $-\infty$ .

STEP 3

Analyze the behavior of the numerator $x-4$ as $x$ approaches $-\infty$ .
As $x \rightarrow -\infty$ , the term $x-4$ tends to $-\infty$ .

STEP 4

Analyze the behavior of the denominator $x^2 + 1$ as $x$ approaches $-\infty$ .
As $x \rightarrow -\infty$ , the term $x^2 + 1$ tends to $\infty$ .

STEP 5

Simplify the expression $\frac{x-4}{x^2+1}$ by dividing both the numerator and the denominator by $x^2$ .
$\frac{x-4}{x^2+1} = \frac{x/x^2 - 4/x^2}{x^2/x^2 + 1/x^2} = \frac{\frac{1}{x} - \frac{4}{x^2}}{1 + \frac{1}{x^2}}$

STEP 6

Evaluate the limit of the simplified expression as $x$ approaches $-\infty$ .
As $x \rightarrow -\infty$ , $\frac{1}{x} \rightarrow 0 \quad \text{and} \quad \frac{4}{x^2} \rightarrow 0,$ so the expression simplifies to $\frac{0 - 0}{1 + 0} = 0.$ Therefore, $\lim _{x \rightarrow -\infty} \frac{x-4}{x^2+1} = 0.$

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