To find all the asymptotes of the function $f(x) = x^2 \left( e^{\frac{1}{x}} - 1 \right)$, we need to examine the behavior of the function as $x$ approaches specific values, usually $x \to \infty$, $x \to -\infty$, and near any points where the function might become undefined (e.g., vertical asymptotes).

Step 1: Investigate as $x \to \infty$

As $x \to \infty$, we have:

$\frac{1}{x} \to 0$

Thus, $e^{\frac{1}{x}} \to e^0 = 1$, and so:

$f(x) = x^2 (e^{\frac{1}{x}} - 1) \to x^2 (1 - 1) = 0$

Therefore, the function tends to 0 as $x \to \infty$, indicating that the horizontal asymptote is:

y = 0 \quad \text{(as x \to \infty)}

Step 2: Investigate as $x \to -\infty$

As $x \to -\infty$, similarly, $\frac{1}{x} \to 0$ and $e^{\frac{1}{x}} \to e^0 = 1$. So:

$f(x) = x^2 (e^{\frac{1}{x}} - 1) \to x^2 (1 - 1) = 0$

Thus, there is also a horizontal asymptote at:

y = 0 \quad \text{(as x \to -\infty)}

Step 3: Investigate as $x \to 0^+$ (from the right side of 0)

As $x \to 0^+$, $\frac{1}{x} \to \infty$, so $e^{\frac{1}{x}} \to \infty$. Thus:

$f(x) = x^2 \left( e^{\frac{1}{x}} - 1 \right) \sim x^2 \cdot e^{\frac{1}{x}}$

Since $e^{\frac{1}{x}} \to \infty$ much faster than $x^2$ decreases, $f(x) \to \infty$ as $x \to 0^+$. Therefore, there is no vertical asymptote as $x \to 0^+$, but the function diverges to infinity.

Step 4: Investigate as $x \to 0^-$ (from the left side of 0)

As $x \to 0^-$, $\frac{1}{x} \to -\infty$, so $e^{\frac{1}{x}} \to 0$. Therefore:

$f(x) = x^2 \left( e^{\frac{1}{x}} - 1 \right) \sim x^2 \cdot (-1) = -x^2$

Thus, as $x \to 0^-$, $f(x) \to 0$.

Conclusion:

• Horizontal asymptote: $y = 0$ (both as $x \to \infty$ and $x \to -\infty$).
• No vertical asymptote at $x = 0$, but the function diverges as $x \to 0^+$.

Would you like more details or have any questions?

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Here are 5 related questions:

1. How do you find the derivative of $f(x) = x^2(e^{1/x} - 1)$?
2. What are the critical points of the function?
3. Can you determine the concavity of the function at different points?
4. What happens to the function when $x$ approaches zero from both sides?
5. Can you explore limits of other functions with $e^{1/x}$ terms?

Tip: Always check both sides of a function near potential vertical asymptotes for more precise behavior.