Math / Calculus

QuestionFind the limit, if it exists. (If an answer does not exist, enter DNE.) $\lim _{x \rightarrow \infty} \frac{\sqrt{x+5 x^{2}}}{3 x-1}$

Studdy Solution

STEP 1

1. We are dealing with a limit as $x$ approaches infinity.
2. The expression involves both polynomial and square root terms.
3. We will simplify the expression to find the limit more easily.

STEP 2

1. Simplify the expression inside the limit.
2. Factor out the highest power of $x$ from both the numerator and the denominator.
3. Evaluate the limit of the simplified expression as $x$ approaches infinity.

STEP 3

Simplify the expression inside the limit by factoring out the highest power of $x$ from the numerator.
$\frac{\sqrt{x + 5x^2}}{3x - 1}$
Rewrite the square root term by factoring out $x^2$ from inside the square root.
$\frac{\sqrt{x(1/x + 5x)}}{3x - 1} = \frac{\sqrt{x \cdot 5x (1/5x + 1)}}{3x - 1} = \frac{\sqrt{5x^2(1/5 + 1/x)}}{3x - 1}$

STEP 4

Simplify the square root expression further.
$\frac{\sqrt{5x^2 (1/5 + 1/x)}}{3x - 1} = \frac{\sqrt{5x^2} \cdot \sqrt{1/5 + 1/x}}{3x - 1} = \frac{\sqrt{5} x \cdot \sqrt{1/5 + 1/x}}{3x - 1}$

STEP 5

Divide the numerator and the denominator by $x$ to simplify the expression.
$\frac{\sqrt{5} x \cdot \sqrt{1/5 + 1/x}}{3x - 1} = \frac{\sqrt{5} x \cdot \sqrt{1/5 + 1/x}}{x (3 - 1/x)}$

STEP 6

Simplify the expression by canceling out $x$ in the numerator and the denominator.
$\frac{\sqrt{5} \cdot \sqrt{1/5 + 1/x}}{3 - 1/x}$

STEP 7

Evaluate the limit as $x$ approaches infinity. Note that $1/x$ approaches 0 as $x$ approaches infinity.
$\lim_{x \to \infty} \frac{\sqrt{5} \cdot \sqrt{1/5 + 1/x}}{3 - 1/x} = \frac{\sqrt{5} \cdot \sqrt{1/5 + 0}}{3 - 0} = \frac{\sqrt{5} \cdot \sqrt{1/5}}{3}$

STEP 8

Simplify the final expression.
$\frac{\sqrt{5} \cdot \sqrt{1/5}}{3} = \frac{\sqrt{5} \cdot \frac{1}{\sqrt{5}}}{3} = \frac{1}{3}$
Therefore, the limit is:
$\lim_{x \to \infty} \frac{\sqrt{x + 5x^2}}{3x - 1} = \frac{1}{3}$

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