## Calculus: Early Transcendentals 8th Edition

$$\lim_{u\to\infty}\frac{e^{u/10}}{u^3}=\infty$$
$$A=\lim_{u\to\infty}\frac{e^{u/10}}{u^3}$$ As $u\to\infty$, $e^{u/10}\to\infty$ and $u^3\to\infty$. That means we have an indeterminate form of $\infty/\infty$ here, favourable to the application of L'Hospital's Rule: $$A=\lim_{u\to\infty}\frac{(e^{u/10})'}{(u^3)'}$$ $$A=\lim_{u\to\infty}\frac{e^{u/10}(u/10)'}{3u^2}$$ $$A=\lim_{u\to\infty}\frac{e^{u/10}}{30u^2}$$ As $u\to\infty$, $e^{u/10}\to\infty$ and $u^2\to\infty$. So again, we have an indeterminate form of $\infty/\infty$. We apply L'Hospital's Rule one more time: $$A=\lim_{u\to\infty}\frac{e^{u/10}(u/10)'}{60u}$$ $$A=\lim_{u\to\infty}\frac{e^{u/10}}{600u}$$ Again, as $u\to\infty$, $e^{u/10}\to\infty$ and $u\to\infty$. Another indeterminate form of $\infty/\infty$, meaning we still can apply L'Hospital's Rule: $$A=\lim_{u\to\infty}\frac{e^{u/10}(u/10)'}{600}$$ $$A=\lim_{u\to\infty}\frac{e^{u/10}}{6000}$$ As $u\to\infty$, $e^{u/10}\to\infty$, so $\frac{e^{u/10}}{6000}$ also approaches $\infty.$ In other words, $$A=\infty$$