Question

Theory and ApplicationsL’Hôpital’s Rule does not help with the limits in Exercises 69–76. Try it—you just keep on cycling. Find the limits some other way.75. lim (x → ∞) e^(x²) / (x e^x)

Accepted Answer

First, identify the form of the limit as $$x 	o \infty$$ for the expression $$\frac{e$$^{x^{2}$$}}{x e$$^{x}$$}. $$Notice that the numerator grows like $$e^{x^{2}$$}$$ and the denominator grows like x$$ $$e^{x}. $$Recognize that $$e^{x^{2}$$}$$ grows much faster than e$$^{x}$$ as x$ becomes very large, since the exponent x$$^{2}$$ grows faster than x$. This suggests the limit might be infinite, but we need to confirm rigorously. Since L’Hôpital’s Rule cycles without resolving the limit, try rewriting the expression to compare growth rates more clearly. For example, write the expression as $$\frac{$$e^{x^{2}$$}}{x $$e^{x}$$} = \frac{$$e^{x^{2}$$}}{$$e^{x}$$} \cdot \frac{1}{x} = $$e^{x^{2} - x$$} \cdot \frac{1}{x}$$. Analyze the exponent in the exponential term: x$$^{2}$$ - x. As$$ $$x 	o \infty$$, $$x^{2} - x$ also tends to infinity, so e$$^{x^{2}$$ - x}$$ grows without bound much faster than $$x$$ grows in the denominator. Conclude that since $$e^{x^{2} - x$$}$$ dominates the $$\frac{1}{x}$$ term, the entire expression grows without bound, and thus the limit tends to infinity.

Theory and Applications
L’Hôpital’s Rule does not help with the limits in Exercises 69–76.
Try it—you just keep on cycling. Find the limits some other way.
75. lim (x → ∞) e^(x²) / (x e^x)

Risposta video verificata per un problema simile:

Concetti chiave

Limits at Infinity

Growth Rates of Functions

L’Hôpital’s Rule and Its Limitations