The value of `lim_(xrarroo)(sqrt(n^2+1)+sqrt(n))/((n^4+n)^(1/4)+4sqrt(n))`, is

Options

A. 0
B. 1
C. `-1`
D. none of these

Answer

To solve the limit

\[ \lim_{n \to \infty} \frac{\sqrt{n^2 + 1} + \sqrt{n}}{(n^4 + n)^{1/4} + 4\sqrt{n}}, \]

we will simplify the expression step by step.

### Step 1: Simplify the Numerator
The numerator is \(\sqrt{n^2 + 1} + \sqrt{n}\). We can factor out \(\sqrt{n^2}\) from \(\sqrt{n^2 + 1}\):

\[ \sqrt{n^2 + 1} = \sqrt{n^2(1 + \frac{1}{n^2})} = \sqrt{n^2} \sqrt{1 + \frac{1}{n^2}} = n\sqrt{1 + \frac{1}{n^2}}. \]

Thus, the numerator becomes:

\[ n\sqrt{1 + \frac{1}{n^2}} + \sqrt{n} = n\sqrt{1 + \frac{1}{n^2}} + n^{1/2}. \]

### Step 2: Simplify the Denominator
The denominator is \((n^4 + n)^{1/4} + 4\sqrt{n}\). We can factor out \(n^4\) from \((n^4 + n)^{1/4}\):

\[ (n^4 + n)^{1/4} = (n^4(1 + \frac{1}{n^3}))^{1/4} = n^{4/4} (1 + \frac{1}{n^3})^{1/4} = n (1 + \frac{1}{n^3})^{1/4}. \]

Thus, the denominator becomes:

\[ n(1 + \frac{1}{n^3})^{1/4} + 4\sqrt{n} = n(1 + \frac{1}{n^3})^{1/4} + 4n^{1/2}. \]

### Step 3: Substitute Back into the Limit
Now we can rewrite the limit:

\[ \lim_{n \to \infty} \frac{n\sqrt{1 + \frac{1}{n^2}} + n^{1/2}}{n(1 + \frac{1}{n^3})^{1/4} + 4n^{1/2}}. \]

### Step 4: Divide Numerator and Denominator by \(n\)
To simplify, we divide both the numerator and the denominator by \(n\):

\[ \lim_{n \to \infty} \frac{\sqrt{1 + \frac{1}{n^2}} + \frac{1}{\sqrt{n}}}{(1 + \frac{1}{n^3})^{1/4} + \frac{4}{\sqrt{n}}}. \]

### Step 5: Evaluate the Limit as \(n \to \infty\)
As \(n \to \infty\):

- \(\sqrt{1 + \frac{1}{n^2}} \to 1\),
- \(\frac{1}{\sqrt{n}} \to 0\),
- \((1 + \frac{1}{n^3})^{1/4} \to 1\),
- \(\frac{4}{\sqrt{n}} \to 0\).

Thus, the limit simplifies to:

\[ \frac{1 + 0}{1 + 0} = \frac{1}{1} = 1. \]

### Final Answer
The value of the limit is:

\[ \boxed{1}. \]