Math / Calculus

QuestionFind the limit. (If the limit is infinite, enter ' $\infty$ ' or ' $-\infty$ ', as appropriate. If the limit does not otherwise exist, enter DNE.) $\lim _{x \rightarrow \infty} \frac{\sqrt{x+6 x^{2}}}{2 x-1}$ $\square$ Need Help? Read II Watch it

Studdy Solution

STEP 1

1. The limit involves a rational function with a square root in the numerator and a linear term in the denominator.
2. Both the numerator and the denominator grow as $x$ approaches $\infty$ .
3. Simplifying the expression by factoring out the highest power of $x$ will help in determining the limit as $x \rightarrow \infty$ .

STEP 2

1. Simplify the expression inside the square root in the numerator.
2. Factor out the highest power of $x$ from both the numerator and the denominator.
3. Simplify the resulting expression to evaluate the limit as $x$ approaches $\infty$ .

STEP 3

Simplify the expression inside the square root in the numerator.
$\sqrt{x + 6x^2} = \sqrt{6x^2 + x}$

STEP 4

Factor out the highest power of $x$ , which is $x^2$ , from inside the square root.
$\sqrt{6x^2 + x} = \sqrt{x^2(6 + \frac{1}{x})}$

STEP 5

Simplify the square root expression.
$\sqrt{x^2(6 + \frac{1}{x})} = x\sqrt{6 + \frac{1}{x}}$

STEP 6

Rewrite the original expression with the simplified numerator.
$\frac{\sqrt{x+6x^2}}{2x-1} = \frac{x\sqrt{6 + \frac{1}{x}}}{2x - 1}$

STEP 7

Factor out $x$ from the denominator for easier simplification.
$\frac{x\sqrt{6 + \frac{1}{x}}}{2x - 1} = \frac{x\sqrt{6 + \frac{1}{x}}}{x(2 - \frac{1}{x})}$

STEP 8

Simplify the expression by canceling out $x$ in the numerator and denominator.
$\frac{x\sqrt{6 + \frac{1}{x}}}{x(2 - \frac{1}{x})} = \frac{\sqrt{6 + \frac{1}{x}}}{2 - \frac{1}{x}}$

STEP 9

Evaluate the limit as $x \rightarrow \infty$ . As $x$ approaches $\infty$ , $\frac{1}{x}$ approaches $0$ .
$\lim_{x \to \infty} \frac{\sqrt{6 + \frac{1}{x}}}{2 - \frac{1}{x}} = \frac{\sqrt{6 + 0}}{2 - 0} = \frac{\sqrt{6}}{2}$
Solution: $\frac{\sqrt{6}}{2}$

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QuestionFind the limit. (If the limit is infinite, enter ' $\infty$ ' or ' $-\infty$ ', as appropriate. If the limit does not otherwise exist, enter DNE.) $\lim _{x \rightarrow \infty} \frac{\sqrt{x+6 x^{2}}}{2 x-1}$ $\square$ Need Help? Read II Watch it

Studdy Solution

STEP 1

STEP 2

1. Simplify the expression inside the square root in the numerator.
2. Factor out the highest power of $x$ from both the numerator and the denominator.
3. Simplify the resulting expression to evaluate the limit as $x$ approaches $\infty$ .

STEP 3

Simplify the expression inside the square root in the numerator.
$\sqrt{x + 6x^2} = \sqrt{6x^2 + x}$

STEP 4

Factor out the highest power of $x$ , which is $x^2$ , from inside the square root.
$\sqrt{6x^2 + x} = \sqrt{x^2(6 + \frac{1}{x})}$

STEP 5

Simplify the square root expression.
$\sqrt{x^2(6 + \frac{1}{x})} = x\sqrt{6 + \frac{1}{x}}$

STEP 6

Rewrite the original expression with the simplified numerator.
$\frac{\sqrt{x+6x^2}}{2x-1} = \frac{x\sqrt{6 + \frac{1}{x}}}{2x - 1}$

STEP 7

Factor out $x$ from the denominator for easier simplification.
$\frac{x\sqrt{6 + \frac{1}{x}}}{2x - 1} = \frac{x\sqrt{6 + \frac{1}{x}}}{x(2 - \frac{1}{x})}$

STEP 8

Simplify the expression by canceling out $x$ in the numerator and denominator.
$\frac{x\sqrt{6 + \frac{1}{x}}}{x(2 - \frac{1}{x})} = \frac{\sqrt{6 + \frac{1}{x}}}{2 - \frac{1}{x}}$

STEP 9

Evaluate the limit as $x \rightarrow \infty$ . As $x$ approaches $\infty$ , $\frac{1}{x}$ approaches $0$ .
$\lim_{x \to \infty} \frac{\sqrt{6 + \frac{1}{x}}}{2 - \frac{1}{x}} = \frac{\sqrt{6 + 0}}{2 - 0} = \frac{\sqrt{6}}{2}$
Solution: $\frac{\sqrt{6}}{2}$

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Studdy Solution

STEP 1

STEP 2

1. Simplify the expression inside the square root in the numerator.2. Factor out the highest power of x x x from both the numerator and the denominator.3. Simplify the resulting expression to evaluate the limit as x x x approaches ∞\infty∞.

STEP 3

Simplify the expression inside the square root in the numerator.x+6x2=6x2+x \sqrt{x + 6x^2} = \sqrt{6x^2 + x} x+6x2​=6x2+x​

STEP 4

Factor out the highest power of x x x, which is x2 x^2 x2, from inside the square root.6x2+x=x2(6+1x) \sqrt{6x^2 + x} = \sqrt{x^2(6 + \frac{1}{x})} 6x2+x​=x2(6+x1​)​

STEP 5

Simplify the square root expression.x2(6+1x)=x6+1x \sqrt{x^2(6 + \frac{1}{x})} = x\sqrt{6 + \frac{1}{x}} x2(6+x1​)​=x6+x1​​

STEP 6

Rewrite the original expression with the simplified numerator.x+6x22x−1=x6+1x2x−1 \frac{\sqrt{x+6x^2}}{2x-1} = \frac{x\sqrt{6 + \frac{1}{x}}}{2x - 1} 2x−1x+6x2​​=2x−1x6+x1​​​

STEP 7

Factor out x x x from the denominator for easier simplification.x6+1x2x−1=x6+1xx(2−1x) \frac{x\sqrt{6 + \frac{1}{x}}}{2x - 1} = \frac{x\sqrt{6 + \frac{1}{x}}}{x(2 - \frac{1}{x})} 2x−1x6+x1​​​=x(2−x1​)x6+x1​​​

STEP 8

Simplify the expression by canceling out x x x in the numerator and denominator.x6+1xx(2−1x)=6+1x2−1x \frac{x\sqrt{6 + \frac{1}{x}}}{x(2 - \frac{1}{x})} = \frac{\sqrt{6 + \frac{1}{x}}}{2 - \frac{1}{x}} x(2−x1​)x6+x1​​​=2−x1​6+x1​​​

STEP 9

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1. Simplify the expression inside the square root in the numerator.
2. Factor out the highest power of $x$ from both the numerator and the denominator.
3. Simplify the resulting expression to evaluate the limit as $x$ approaches $\infty$ .

Simplify the expression inside the square root in the numerator.
$\sqrt{x + 6x^2} = \sqrt{6x^2 + x}$

Factor out the highest power of $x$ , which is $x^2$ , from inside the square root.
$\sqrt{6x^2 + x} = \sqrt{x^2(6 + \frac{1}{x})}$

Simplify the square root expression.
$\sqrt{x^2(6 + \frac{1}{x})} = x\sqrt{6 + \frac{1}{x}}$

Rewrite the original expression with the simplified numerator.
$\frac{\sqrt{x+6x^2}}{2x-1} = \frac{x\sqrt{6 + \frac{1}{x}}}{2x - 1}$

Factor out $x$ from the denominator for easier simplification.
$\frac{x\sqrt{6 + \frac{1}{x}}}{2x - 1} = \frac{x\sqrt{6 + \frac{1}{x}}}{x(2 - \frac{1}{x})}$

Simplify the expression by canceling out $x$ in the numerator and denominator.
$\frac{x\sqrt{6 + \frac{1}{x}}}{x(2 - \frac{1}{x})} = \frac{\sqrt{6 + \frac{1}{x}}}{2 - \frac{1}{x}}$