Answer :
To find the limit [tex]\(\lim_{x \to \infty} \frac{\frac{1}{x} - 10x}{2x + \frac{1}{3x}}\)[/tex], let's go through the steps in detail.
Given:
[tex]\[ f(x) = \frac{\frac{1}{x} - 10x}{2x + \frac{1}{3x}} \][/tex]
1. Simplify the expression:
First, we need to break down the given fraction. Let's look at the numerator and the denominator separately:
[tex]\[ \text{Numerator}: \frac{1}{x} - 10x \][/tex]
[tex]\[ \text{Denominator}: 2x + \frac{1}{3x} \][/tex]
2. Factor out the dominant terms:
When [tex]\(x\)[/tex] approaches infinity, the dominant term in the numerator [tex]\(\frac{1}{x} - 10x\)[/tex] is [tex]\(-10x\)[/tex] because [tex]\( \frac{1}{x} \)[/tex] approaches 0.
Similarly, the dominant term in the denominator [tex]\(2x + \frac{1}{3x}\)[/tex] is [tex]\(2x\)[/tex] because [tex]\(\frac{1}{3x}\)[/tex] approaches 0.
3. Divide the numerator and the denominator by [tex]\(x\)[/tex]:
Let's rewrite the function by dividing both the numerator and denominator by [tex]\(x\)[/tex]:
[tex]\[ \frac{\frac{\frac{1}{x} - 10x}{x}}{\frac{2x + \frac{1}{3x}}{x}} = \frac{\frac{1}{x^2} - 10}{2 + \frac{1}{3x^2}} \][/tex]
4. Simplify the terms:
As [tex]\(x \to \infty\)[/tex], [tex]\(\frac{1}{x^2} \to 0\)[/tex] and [tex]\(\frac{1}{3x^2} \to 0\)[/tex]. Therefore, the expression simplifies to:
[tex]\[ \frac{0 - 10}{2 + 0} = \frac{-10}{2} = -5 \][/tex]
Thus, the limit as [tex]\( x \)[/tex] approaches infinity of the given function is:
[tex]\[ \lim_{x \to \infty} \frac{\frac{1}{x} - 10x}{2x + \frac{1}{3x}} = -5 \][/tex]
Therefore, the answer is [tex]\(-5\)[/tex].
Given:
[tex]\[ f(x) = \frac{\frac{1}{x} - 10x}{2x + \frac{1}{3x}} \][/tex]
1. Simplify the expression:
First, we need to break down the given fraction. Let's look at the numerator and the denominator separately:
[tex]\[ \text{Numerator}: \frac{1}{x} - 10x \][/tex]
[tex]\[ \text{Denominator}: 2x + \frac{1}{3x} \][/tex]
2. Factor out the dominant terms:
When [tex]\(x\)[/tex] approaches infinity, the dominant term in the numerator [tex]\(\frac{1}{x} - 10x\)[/tex] is [tex]\(-10x\)[/tex] because [tex]\( \frac{1}{x} \)[/tex] approaches 0.
Similarly, the dominant term in the denominator [tex]\(2x + \frac{1}{3x}\)[/tex] is [tex]\(2x\)[/tex] because [tex]\(\frac{1}{3x}\)[/tex] approaches 0.
3. Divide the numerator and the denominator by [tex]\(x\)[/tex]:
Let's rewrite the function by dividing both the numerator and denominator by [tex]\(x\)[/tex]:
[tex]\[ \frac{\frac{\frac{1}{x} - 10x}{x}}{\frac{2x + \frac{1}{3x}}{x}} = \frac{\frac{1}{x^2} - 10}{2 + \frac{1}{3x^2}} \][/tex]
4. Simplify the terms:
As [tex]\(x \to \infty\)[/tex], [tex]\(\frac{1}{x^2} \to 0\)[/tex] and [tex]\(\frac{1}{3x^2} \to 0\)[/tex]. Therefore, the expression simplifies to:
[tex]\[ \frac{0 - 10}{2 + 0} = \frac{-10}{2} = -5 \][/tex]
Thus, the limit as [tex]\( x \)[/tex] approaches infinity of the given function is:
[tex]\[ \lim_{x \to \infty} \frac{\frac{1}{x} - 10x}{2x + \frac{1}{3x}} = -5 \][/tex]
Therefore, the answer is [tex]\(-5\)[/tex].
