The value of `lim_(xrarroo) (e^(sqrt(x^(4)+))-e^((x^(2)+1)))` is

Options

A. 0
B. e
C. 1/e
D. `-oo`

Explanation

To solve the limit \( \lim_{x \to \infty} \left( e^{\sqrt{x^4 + 1}} - e^{x^2 + 1} \right) \), we will follow these steps:

### Step 1: Analyze the expression inside the limit
We start with the expression:
\[ e^{\sqrt{x^4 + 1}} - e^{x^2 + 1} \]
As \( x \to \infty \), we need to analyze the behavior of both terms.

### Step 2: Simplify \( \sqrt{x^4 + 1} \)
For large \( x \):
\[ \sqrt{x^4 + 1} \approx \sqrt{x^4} = x^2 \]
Thus,
\[ \sqrt{x^4 + 1} = x^2 \sqrt{1 + \frac{1}{x^4}} \approx x^2 \left(1 + \frac{1}{2x^4}\right) = x^2 + \frac{1}{2x^2} \]
So,
\[ e^{\sqrt{x^4 + 1}} \approx e^{x^2 + \frac{1}{2x^2}} = e^{x^2} e^{\frac{1}{2x^2}} \]

### Step 3: Expand \( e^{\frac{1}{2x^2}} \)
Using the Taylor expansion for \( e^u \) where \( u = \frac{1}{2x^2} \):
\[ e^{\frac{1}{2x^2}} \approx 1 + \frac{1}{2x^2} + O\left(\frac{1}{x^4}\right) \]
Thus,
\[ e^{\sqrt{x^4 + 1}} \approx e^{x^2} \left(1 + \frac{1}{2x^2}\right) \approx e^{x^2} + \frac{e^{x^2}}{2x^2} \]

### Step 4: Analyze \( e^{x^2 + 1} \)
We can rewrite \( e^{x^2 + 1} \) as:
\[ e^{x^2 + 1} = e^{x^2} e \]

### Step 5: Combine the two expressions
Now we can substitute back into our limit:
\[ e^{\sqrt{x^4 + 1}} - e^{x^2 + 1} \approx \left(e^{x^2} + \frac{e^{x^2}}{2x^2}\right) - e \cdot e^{x^2} \]
This simplifies to:
\[ e^{x^2} \left(1 + \frac{1}{2x^2} - e\right) \]

### Step 6: Factor out \( e^{x^2} \)
Thus, we have:
\[ e^{x^2} \left(\frac{1}{2x^2} + 1 - e\right) \]

### Step 7: Evaluate the limit
As \( x \to \infty \), \( e^{x^2} \) grows exponentially, and the term \( \left(\frac{1}{2x^2} + 1 - e\right) \) approaches \( 1 - e \). Therefore:
\[ \lim_{x \to \infty} \left( e^{\sqrt{x^4 + 1}} - e^{x^2 + 1} \right) = \infty \cdot (1 - e) \]
Since \( e > 1 \), \( 1 - e < 0 \), thus:
\[ \lim_{x \to \infty} \left( e^{\sqrt{x^4 + 1}} - e^{x^2 + 1} \right) = -\infty \]

### Final Answer:
The value of the limit is:
\[ \boxed{-\infty} \]