To determine the limit of the sequence [tex]\( b_n = n^{4/n} \)[/tex] as [tex]\( n \)[/tex] approaches infinity, we can use properties of exponential and logarithmic functions. Let’s proceed step-by-step:
### Step 1: Rewrite the expression using logarithms
Consider the sequence [tex]\( b_n = n^{4/n} \)[/tex]. To simplify the limit computation, take the natural logarithm of both sides:
[tex]\[
\ln(b_n) = \ln\left(n^{4/n}\right)
\][/tex]
Using the property of logarithms that [tex]\( \ln(a^b) = b \ln(a) \)[/tex], we can rewrite this as:
[tex]\[
\ln(b_n) = \frac{4}{n} \ln(n)
\][/tex]
Now, let's define a new sequence [tex]\( L_n \)[/tex]:
[tex]\[
L_n = \frac{4 \ln(n)}{n}
\][/tex]
### Step 2: Determine the limit of [tex]\( \ln(b_n) \)[/tex]
We need to find the limit of [tex]\( L_n \)[/tex] as [tex]\( n \)[/tex] approaches infinity:
[tex]\[
\lim_{n \to \infty} L_n = \lim_{n \to \infty} \frac{4 \ln(n)}{n}
\][/tex]
### Step 3: Apply L'Hôpital's Rule
The limit [tex]\( \lim_{n \to \infty} \frac{\ln(n)}{n} \)[/tex] is of the form [tex]\( \frac{\infty}{\infty} \)[/tex]. To resolve this indeterminate form, apply L'Hôpital's Rule:
[tex]\[
\lim_{n \to \infty} \frac{\ln(n)}{n} = \lim_{n \to \infty} \frac{\frac{d}{dn}(\ln(n))}{\frac{d}{dn}(n)} = \lim_{n \to \infty} \frac{1/n}{1} = \lim_{n \to \infty} \frac{1}{n} = 0
\][/tex]
Since [tex]\( \lim_{n \to \infty} \frac{\ln(n)}{n} = 0 \)[/tex], we have:
[tex]\[
\lim_{n \to \infty} L_n = \lim_{n \to \infty} \frac{4 \ln(n)}{n} = 4 \cdot \lim_{n \to \infty} \frac{\ln(n)}{n} = 4 \cdot 0 = 0
\][/tex]
### Step 4: Convert back from the logarithmic limit to the original sequence
We found that [tex]\( \lim_{n \to \infty} \ln(b_n) = 0 \)[/tex]. Recall that if [tex]\( \ln(a_n) \to L \)[/tex], then [tex]\( a_n = e^L \)[/tex]. Hence,
[tex]\[
\lim_{n \to \infty} b_n = e^0 = 1
\][/tex]
### Conclusion
Therefore, the limit of the given sequence as [tex]\( n \)[/tex] approaches infinity is:
[tex]\[
\lim_{n \rightarrow \infty} b_n = 1
\][/tex]
