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Evaluate the limit:

[tex]\[ \lim_{x \rightarrow \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}} \][/tex]

Sagot :

GinnyAnswer
To find the limit \(\lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}\), let's analyze the expression step by step.

### Step 1: Factor and Simplify
Consider the expression:
[tex]\[ \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}} \][/tex]

First, let's rewrite \(\sqrt[3]{x}\) as \(x^{1/3}\) and \(\sqrt[5]{x}\) as \(x^{1/5}\). The expression becomes:
[tex]\[ \frac{x^{1/3} - x^{1/5}}{x^{1/3} + x^{1/5}} \][/tex]

### Step 2: Factor Out the Dominant Term
The dominant term in both the numerator and denominator as \(x \to \infty\) is \(x^{1/3}\). Factor \(x^{1/3}\) out from both the numerator and the denominator:
[tex]\[ \frac{x^{1/3} (1 - x^{(1/5 - 1/3)})}{x^{1/3} (1 + x^{(1/5 - 1/3)})} \][/tex]

Notice that \(1/5\) is less than \(1/3\), so \(1/5 - 1/3\) is negative. Let’s simplify the exponent:
[tex]\[ \frac{x^{1/3} (1 - x^{-2/15})}{x^{1/3} (1 + x^{-2/15})} \][/tex]

### Step 3: Cancelling Out x^{1/3}
Now, cancel the common factor of \(x^{1/3}\) from both the numerator and the denominator:
[tex]\[ \frac{1 - x^{-2/15}}{1 + x^{-2/15}} \][/tex]

### Step 4: Evaluate the Limit
As \(x \to \infty\), \(x^{-2/15} \to 0\). Thus, the expression simplifies to:
[tex]\[ \frac{1 - 0}{1 + 0} = \frac{1}{1} = 1 \][/tex]

### Final Answer
Therefore, the limit is:
[tex]\[ \lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}} = 1 \][/tex]
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