Use the appropriate limit laws and theorems to determine the limit of the sequence.

[tex]\[ b_n = n^{4 / n} \][/tex]

(Use symbolic notation and fractions where needed. Enter DNE if the sequence diverges.)

[tex]\[ \lim_{n \rightarrow \infty} b_n = \boxed{\square} \][/tex]



Answer :

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]