Questions: Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. lim as x approaches infinity of sqrt(x) * e^(-x / 2)

Solution Steps

To find the limit of \(\lim _{x \rightarrow \infty} \sqrt{x} e^{-x / 2}\), we can analyze the behavior of the function as \(x\) approaches infinity. The function \(\sqrt{x}\) grows without bound, but \(e^{-x/2}\) decays exponentially to zero. The exponential decay is much faster than the polynomial growth, suggesting the limit is zero. We can confirm this by applying l'Hospital's Rule, which is applicable here since the expression is of the indeterminate form \(\frac{\infty}{\infty}\).

1. Rewrite the expression as a fraction: \(\frac{\sqrt{x}}{e^{x/2}}\).
2. Apply l'Hospital's Rule, which involves differentiating the numerator and the denominator until the limit can be evaluated directly.
3. Evaluate the limit of the resulting expression as \(x\) approaches infinity.

We start with the limit we want to evaluate: \[ \lim _{x \rightarrow \infty} \sqrt{x} e^{-x / 2} \] This can be rewritten as a fraction: \[ \lim _{x \rightarrow \infty} \frac{\sqrt{x}}{e^{x/2}} \]

Since both the numerator \(\sqrt{x}\) and the denominator \(e^{x/2}\) approach infinity as \(x\) approaches infinity, we can apply l'Hospital's Rule. We differentiate the numerator and the denominator:

• The derivative of \(\sqrt{x}\) is \(\frac{1}{2\sqrt{x}}\).
• The derivative of \(e^{x/2}\) is \(\frac{1}{2} e^{x/2}\).

Thus, we have: \[ \lim _{x \rightarrow \infty} \frac{\frac{1}{2\sqrt{x}}}{\frac{1}{2} e^{x/2}} = \lim _{x \rightarrow \infty} \frac{1}{\sqrt{x} e^{x/2}} \]

As \(x\) approaches infinity, \(\sqrt{x}\) grows without bound while \(e^{x/2}\) grows exponentially. Therefore, the limit simplifies to: \[ \lim _{x \rightarrow \infty} \frac{1}{\sqrt{x} e^{x/2}} = 0 \]

Final Answer