Two Limits by L'Hospital: e^x / x^2 and x^2 ln x
Evaluate the following limits, justifying each step.
(a) $\displaystyle \lim_{x \to \infty} \frac{e^{x}}{x^{2}}$
(b) $\displaystyle \lim_{x \to 0^{+}} x^{2} \ln x$
Hints
- Part (a) is an $\infty/\infty$ form and part (b) is a $0 \cdot (-\infty)$ form. L'Hospital's rule only applies to quotients of type $0/0$ or $\infty/\infty$, so rewrite (b) as a quotient first.
- For (a), apply L'Hospital twice: the numerator never changes but the denominator's degree drops by one each time.
- For (b), write $x^2 \ln x = \dfrac{\ln x}{x^{-2}}$, differentiate top and bottom, and simplify before taking the limit.
Worked Solution
How to Think About It: Both limits are indeterminate forms. L'Hospital's rule handles $0/0$ and $\infty/\infty$ quotients, so the only work is to put each expression in that shape and differentiate until the indeterminacy disappears.
Quick Estimate: At $x = 20$, $e^{20} \approx 4.9 \times 10^{8}$ while $x^2 = 400$, so the ratio is already about $10^{6}$ and growing. At $x = 0.01$, $x^2 \ln x = 10^{-4} \times (-4.6) \approx -0.00046$, clearly heading to $0$.
Formal Solution:
Part (a): $\lim_{x \to \infty} e^{x}/x^{2}$
*Step 1 -- Check the form.* As $x \to \infty$ both $e^x$ and $x^2$ tend to $\infty$, so the quotient is of type $\infty/\infty$ and L'Hospital applies.
*Step 2 -- Apply L'Hospital twice.* $$\lim_{x \to \infty} \frac{e^{x}}{x^{2}} = \lim_{x \to \infty} \frac{e^{x}}{2x} = \lim_{x \to \infty} \frac{e^{x}}{2} = \infty.$$ After the first application the form is still $\infty/\infty$, so a second application is legitimate; the second one leaves a quotient with a constant denominator.
Part (b): $\lim_{x \to 0^{+}} x^{2} \ln x$
*Step 1 -- Rewrite as a quotient.* As $x \to 0^{+}$, $x^2 \to 0$ and $\ln x \to -\infty$: a $0 \cdot \infty$ form. Write $$x^{2} \ln x = \frac{\ln x}{x^{-2}},$$ which is of type $-\infty/\infty$.
*Step 2 -- Apply L'Hospital once.* $$\lim_{x \to 0^{+}} \frac{\ln x}{x^{-2}} = \lim_{x \to 0^{+}} \frac{1/x}{-2x^{-3}} = \lim_{x \to 0^{+}} \left(-\frac{x^{2}}{2}\right) = 0.$$
*Step 3 -- Note the general pattern.* The same computation gives $\lim_{x \to 0^{+}} x^{a} \ln x = 0$ for every $a > 0$, and $\lim_{x \to \infty} e^{x}/x^{k} = \infty$ for every fixed $k$.
Answer: (a) $+\infty$ (the exponential dominates any polynomial). (b) $0$ (the power $x^2$ dominates the logarithm near $0$).
Intuition
Exponentials dominate every polynomial at infinity, and every positive power of $x$ crushes $\ln x$ near zero: $e^x/x^k \to \infty$ and $x^a \ln x \to 0$ for any $a > 0$. These two hierarchies are the workhorse facts behind tail estimates (Gaussian tails beat polynomial decay), entropy calculations where $p \ln p \to 0$ as $p \to 0$, and any asymptotic argument in complexity or convergence-rate analysis.