Find lim as x to infinity (e^x)/(x^5+4x²+5)
Find $\displaystyle\lim_{x\to\infty}\frac{e^{x}}{x^{5}+4x^{2}+5}$.
Explanation of Method
Five differentiations exhaust the polynomial; the exponential survives.
Step 1: The form
Both parts tend to $+\infty$, so the form is $\dfrac{\infty}{\infty}$. The competition is between an exponential and a polynomial of degree $5$.
Step 2: Apply l'Hôpital's rule five times
Each application leaves the numerator unchanged and lowers the degree of the denominator by one: \[\frac{e^{x}}{x^{5}+4x^{2}+5} \to \frac{e^{x}}{5x^{4}+8x} \to \frac{e^{x}}{20x^{3}+8} \to \frac{e^{x}}{60x^{2}} \to \frac{e^{x}}{120x} \to \frac{e^{x}}{120}.\]
Each step is legitimate because every intermediate form is again $\dfrac{\infty}{\infty}$.
Step 3: The last limit is immediate
\[ \lim_{x\to\infty}\frac{e^{x}}{120} = +\infty. \]
Step 4: Verification
\[x = 20:\ \frac{4.85\cdot10^{8}}{3.2\cdot10^{6}} \approx 152; \qquad x = 40:\ \frac{2.35\cdot10^{17}}{1.02\cdot10^{8}} \approx 2.3\cdot10^{9}\]
\[\text{The ratio is still below } 1 \text{ until } x \approx 12, \text{ then grows without bound} \;\;\checkmark\]
\[\text{Equivalently } \frac{x^{5}+4x^{2}+5}{e^{x}} \to 0, \text{ the reciprocal statement of Problem 378} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\lim_{x\to\infty}\frac{e^{x}}{x^{5}+4x^{2}+5} = +\infty\,}\]
Remark
Remark. The moral is that the degree of the polynomial is irrelevant: whatever it is, that many differentiations reduce it to a constant while $e^{x}$ is untouched. The exponential outgrows every polynomial — though, as the numbers show, only after a long delay when the degree is high.