Question

The value of $\lim_{x\to\infty} e^{-x} \int_{\frac{1}{x}}^{x+\frac{1}{x}} g(t) \, dt$ equal to :

• A. 0
• B. 1
• C. 2
• D. DNE

Answer 1

Answer: (B)

1

Answer 2

Let the limit be $L$. We have $L = \lim_{x\to\infty} \frac{\int_{\frac{1}{x}}^{x+\frac{1}{x}} g(t) \, dt}{e^x}$. This is of the indeterminate form $\frac{\infty}{\infty}$ if $\int g(t) dt$ tends to infinity, or $\frac{L'}{\infty}$ if $\int g(t) dt$ converges to $L'$. Consider the case $g(t) = e^t$. Then, $\int_{\frac{1}{x}}^{x+\frac{1}{x}} e^t \, dt = [e^t]_{\frac{1}{x}}^{x+\frac{1}{x}} = e^{x+\frac{1}{x}} - e^{\frac{1}{x}}$. The limit becomes: $L = \lim_{x\to\infty} e^{-x} (e^{x+\frac{1}{x}} - e^{\frac{1}{x}}) = \lim_{x\to\infty} (e^{x+\frac{1}{x}}e^{-x} - e^{\frac{1}{x}}e^{-x})$ $L = \lim_{x\to\infty} (e^{\frac{1}{x}} - e^{-x+\frac{1}{x}})$ As $x \to \infty$, $\frac{1}{x} \to 0$, so $e^{\frac{1}{x}} \to e^0 = 1$. As $x \to \infty$, $-x+\frac{1}{x} \to -\infty$, so $e^{-x+\frac{1}{x}} \to 0$. Therefore, $L = 1 - 0 = 1$. While the value of the limit depends on the function $g(t)$, in the context of a multiple-choice question with these options, the case $g(t) = e^t$ is often the intended scenario that yields a specific numerical answer among the choices. If $g(t)$ grows slower than $e^t$, the limit is 0. If $g(t)$ grows faster than $e^t$, the limit might be $\infty$ (DNE). The choice 1 suggests that $g(t)$ grows comparably to $e^t$.