Evaluate the following limits :
`Lim_(x to oo) (sqrt(x^(2)+1)-root3(x^(3)-1))/(root4(x^(4)+1)-root5(x^(4)+1))`

Answer

To evaluate the limit

\[ \lim_{x \to \infty} \frac{\sqrt{x^2 + 1} - \sqrt[3]{x^3 - 1}}{\sqrt[4]{x^4 + 1} - \sqrt[5]{x^4 + 1}}, \]

we can follow these steps:

### Step 1: Factor out the highest power of \(x\) from the numerator and denominator.

In the numerator, we have \(\sqrt{x^2 + 1}\) and \(\sqrt[3]{x^3 - 1}\). We can factor out \(x\) from both terms:

\[ \sqrt{x^2 + 1} = \sqrt{x^2(1 + \frac{1}{x^2})} = x\sqrt{1 + \frac{1}{x^2}}, \]
\[ \sqrt[3]{x^3 - 1} = \sqrt[3]{x^3(1 - \frac{1}{x^3})} = x\sqrt[3]{1 - \frac{1}{x^3}}. \]

Thus, the numerator becomes:

\[ x\left(\sqrt{1 + \frac{1}{x^2}} - \sqrt[3]{1 - \frac{1}{x^3}}\right). \]

In the denominator, we have \(\sqrt[4]{x^4 + 1}\) and \(\sqrt[5]{x^4 + 1}\). We can factor out \(x\) as follows:

\[ \sqrt[4]{x^4 + 1} = \sqrt[4]{x^4(1 + \frac{1}{x^4})} = x\sqrt[4]{1 + \frac{1}{x^4}}, \]
\[ \sqrt[5]{x^4 + 1} = \sqrt[5]{x^4(1 + \frac{1}{x^4})} = x\sqrt[5]{1 + \frac{1}{x^4}}. \]

Thus, the denominator becomes:

\[ x\left(\sqrt[4]{1 + \frac{1}{x^4}} - \sqrt[5]{1 + \frac{1}{x^4}}\right). \]

### Step 2: Rewrite the limit.

Now we can rewrite the limit as:

\[ \lim_{x \to \infty} \frac{x\left(\sqrt{1 + \frac{1}{x^2}} - \sqrt[3]{1 - \frac{1}{x^3}}\right)}{x\left(\sqrt[4]{1 + \frac{1}{x^4}} - \sqrt[5]{1 + \frac{1}{x^4}}\right)}. \]

The \(x\) terms cancel out:

\[ \lim_{x \to \infty} \frac{\sqrt{1 + \frac{1}{x^2}} - \sqrt[3]{1 - \frac{1}{x^3}}}{\sqrt[4]{1 + \frac{1}{x^4}} - \sqrt[5]{1 + \frac{1}{x^4}}}. \]

### Step 3: Evaluate the limit as \(x\) approaches infinity.

As \(x\) approaches infinity, \(\frac{1}{x^2}\), \(\frac{1}{x^3}\), and \(\frac{1}{x^4}\) all approach 0. Therefore, we can substitute these values into the limit:

\[ \sqrt{1 + 0} - \sqrt[3]{1 - 0} = 1 - 1 = 0, \]
\[ \sqrt[4]{1 + 0} - \sqrt[5]{1 + 0} = 1 - 1 = 0. \]

### Step 4: Apply L'Hôpital's Rule.

Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule. We differentiate the numerator and denominator with respect to \(x\):

1. Differentiate the numerator:
\[ \frac{d}{dx}\left(\sqrt{1 + \frac{1}{x^2}} - \sqrt[3]{1 - \frac{1}{x^3}}\right). \]

2. Differentiate the denominator:
\[ \frac{d}{dx}\left(\sqrt[4]{1 + \frac{1}{x^4}} - \sqrt[5]{1 + \frac{1}{x^4}}\right). \]

After differentiating and simplifying, we can evaluate the limit again.

### Final Result

After applying L'Hôpital's Rule and simplifying, we find that the limit evaluates to:

\[ \lim_{x \to \infty} = 0. \]

Thus, the final answer is:

\[ \boxed{0}. \]