Limit of ((n+2)/(n+1)) to the power n+3
Evaluate the limit: $$ L = \lim_{n \to \infty} \left( \frac{n+2}{n+1} \right)^{n+3}. $$
Explanation of Method
As $n \to \infty$, the base $\dfrac{n+2}{n+1} \to 1$ while the exponent $n+3 \to \infty$, so this limit has the indeterminate form $1^\infty$. The standard technique is to write the base as $1 + \dfrac{1}{k}$ for some expression $k$ that also tends to infinity, then split the exponent so that one factor matches $k$ exactly. That piece converges to $e$ by the defining limit $\lim_{k \to \infty}\left(1+\frac{1}{k}\right)^k = e$, and the leftover factor is evaluated separately.
- Rewrite the base $\dfrac{n+2}{n+1}$ in the form $1 + \dfrac{1}{n+1}$.
- Split the exponent $n+3$ into $(n+1) + 2$.
- Apply the definition of $e$ to the $(n+1)$-power piece.
- Evaluate the remaining bounded factor and combine the results.
Step 1 · Rewrite the base
Simplify the fraction inside the parentheses:
$$ \frac{n+2}{n+1} = \frac{(n+1)+1}{n+1} = 1 + \frac{1}{n+1}. $$
So the limit expression becomes:
$$ L = \lim_{n \to \infty} \left( 1 + \frac{1}{n+1} \right)^{n+3}. $$
Step 2 · Split the exponent
Write the exponent $n+3$ as $(n+1) + 2$, so that the power can be separated into a product of two factors:
$$ \left( 1 + \frac{1}{n+1} \right)^{n+3} = \left( 1 + \frac{1}{n+1} \right)^{n+1} \cdot \left( 1 + \frac{1}{n+1} \right)^{2}. $$
Therefore:
$$ L = \lim_{n \to \infty} \left( 1 + \frac{1}{n+1} \right)^{n+1} \cdot \lim_{n \to \infty} \left( 1 + \frac{1}{n+1} \right)^{2}. $$
Step 3 · Evaluate each factor
For the first factor, let $k = n+1$. As $n \to \infty$, also $k \to \infty$, so this is exactly the defining limit of $e$:
$$ \lim_{n \to \infty} \left( 1 + \frac{1}{n+1} \right)^{n+1} = \lim_{k \to \infty} \left( 1 + \frac{1}{k} \right)^{k} = e. $$
For the second factor, the base $1 + \dfrac{1}{n+1} \to 1$ as $n \to \infty$, and it is raised to the fixed power $2$, so:
$$ \lim_{n \to \infty} \left( 1 + \frac{1}{n+1} \right)^{2} = 1^2 = 1. $$
Step 4 · Combine the results
Multiplying the two limits together:
$$ L = e \cdot 1 = e. $$
Summary Table
| Piece | Rewritten as | Limit as $n \to \infty$ |
|---|---|---|
| Base | $1 + \dfrac{1}{n+1}$ | $1$ |
| $\left(1+\frac{1}{n+1}\right)^{n+1}$ | defining limit of $e$ | $e$ |
| $\left(1+\frac{1}{n+1}\right)^{2}$ | fixed power of a term $\to 1$ | $1$ |
Additional Notes
- This is the standard trick for $1^\infty$ indeterminate forms: isolate a sub-exponent that matches the reciprocal inside the base exactly, so it reduces to the definition of $e$.
- Any bounded, fixed power (here, squaring) applied to a base tending to $1$ contributes a factor of $1$ and does not affect the final value.
- The same method applies more generally to limits of the form $\lim_{n\to\infty}\left(1+\frac{a}{n+b}\right)^{cn+d}$, which converge to $e^{ac}$.
Result
The value of the limit is: