To solve the limit:

$\lim _{n\rightarrow +\infty }\left(\frac{\left(1+n\right)^{100}}{e^{\sqrt{n}}}\right),$

we analyze the growth rates of the numerator and the denominator.

Step 1: Analyze the numerator

The numerator $(1 + n)^{100}$ grows polynomially with $n$. For large $n$, the term $1 + n$ approximates $n$, so: $(1 + n)^{100} \sim n^{100}.$

Step 2: Analyze the denominator

The denominator $e^{\sqrt{n}}$ grows exponentially with $\sqrt{n}$. For large $n$, exponential growth dominates polynomial growth.

Step 3: Combine the expressions

For large $n$, the limit behaves as: $\frac{(1 + n)^{100}}{e^{\sqrt{n}}} \sim \frac{n^{100}}{e^{\sqrt{n}}}.$

Step 4: Exponential dominance

As $n \to \infty$, the numerator $n^{100}$ grows polynomially, but the denominator $e^{\sqrt{n}}$ grows exponentially. Exponential growth is much faster than polynomial growth, which drives the fraction to $0$.

Final Answer:

$\lim _{n\rightarrow +\infty }\left(\frac{\left(1+n\right)^{100}}{e^{\sqrt{n}}}\right) = 0.$

Would you like further details or a breakdown of any part? 😊

Related Questions:

1. What are the properties of exponential and polynomial growth?
2. How does the substitution technique simplify such limits?
3. What is the general rule for comparing exponential and polynomial terms?
4. How does L'Hôpital's Rule apply to similar problems?
5. How does the growth of $e^{n^p}$ compare to $n^q$ for $p > 0$ and $q > 0$?

Tip:

In limit problems involving competing growth rates, identify whether the dominant growth is exponential, polynomial, or logarithmic.