Evaluate the following limits :
`Lim_(x to oo )( sqrt(x^(2)+5x+4)-sqrt(x^(2)-3x+4))`

Answer

To evaluate the limit

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

we can follow these steps:

### Step 1: Rationalize the expression
We will rationalize the expression by multiplying and dividing by the conjugate:

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

### Step 2: Apply the difference of squares
Using the difference of squares in the numerator:

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

### Step 3: Simplify the numerator
Now, simplify the numerator:

\[ x^2 + 5x + 4 - x^2 + 3x - 4 = 8x. \]

So, we have:

\[ \lim_{x \to \infty} \frac{8x}{\sqrt{x^2 + 5x + 4} + \sqrt{x^2 - 3x + 4}}. \]

### Step 4: Factor out \(x\) from the square roots
Next, we factor \(x^2\) out of the square roots:

\[ \sqrt{x^2 + 5x + 4} = x\sqrt{1 + \frac{5}{x} + \frac{4}{x^2}}, \]
\[ \sqrt{x^2 - 3x + 4} = x\sqrt{1 - \frac{3}{x} + \frac{4}{x^2}}. \]

### Step 5: Substitute back into the limit
Now substitute these back into the limit:

\[ \lim_{x \to \infty} \frac{8x}{x\left(\sqrt{1 + \frac{5}{x} + \frac{4}{x^2}} + \sqrt{1 - \frac{3}{x} + \frac{4}{x^2}}\right)}. \]

### Step 6: Cancel \(x\)
The \(x\) in the numerator and denominator cancels out:

\[ \lim_{x \to \infty} \frac{8}{\sqrt{1 + \frac{5}{x} + \frac{4}{x^2}} + \sqrt{1 - \frac{3}{x} + \frac{4}{x^2}}}. \]

### Step 7: Evaluate the limit
As \(x \to \infty\), \(\frac{5}{x} \to 0\) and \(\frac{4}{x^2} \to 0\):

\[ \lim_{x \to \infty} \frac{8}{\sqrt{1 + 0 + 0} + \sqrt{1 - 0 + 0}} = \frac{8}{1 + 1} = \frac{8}{2} = 4. \]

### Final Answer
Thus, the limit evaluates to:

\[ \boxed{4}. \]