Find the limit if it exists.
$\lim _{x \rightarrow 0}(1+x)^{1 / 6 x}$
A. $e$
B. $e^{1 / 6}$
C. $\frac{1}{e^6}$
D. $e^6$
Answer (1)
Take the natural logarithm of the function: ln L = lim x → 0 6 x l n ( 1 + x ) .
Apply L'Hopital's rule: ln L = lim x → 0 6 1 + x 1 .
Evaluate the limit: ln L = 6 1 .
Exponentiate to find the limit: L = e 1/6 .
The limit is e 1/6 .
Explanation
Problem Analysis We are asked to find the limit of the function ( 1 + x ) 1/6 x as x approaches 0 . This is an indeterminate form of type 1 ∞ , so we can use L'Hopital's rule after taking the natural logarithm, or we can recognize a standard limit.
Taking the Natural Logarithm Let L = lim x → 0 ( 1 + x ) 1/6 x . To evaluate this limit, we can take the natural logarithm of both sides. This gives us ln L = x → 0 lim ln ( 1 + x ) 1/6 x = x → 0 lim 6 x 1 ln ( 1 + x ) = x → 0 lim 6 x ln ( 1 + x ) .
Applying L'Hopital's Rule Now we have an indeterminate form of type 0 0 , so we can apply L'Hopital's rule. Taking the derivative of the numerator and the denominator with respect to x , we get ln L = x → 0 lim 6 1 + x 1 = x → 0 lim 6 ( 1 + x ) 1 .
Evaluating the Limit Now, we can evaluate the limit by direct substitution: ln L = 6 ( 1 + 0 ) 1 = 6 1 .
Solving for L To find L , we exponentiate both sides: L = e l n L = e 1/6 .
Final Answer Therefore, the limit is e 1/6 .
Examples
In financial mathematics, the limit of ( 1 + x ) 1/ x as x approaches 0 is related to continuous compounding of interest. Suppose you invest a principal amount at an annual interest rate, compounded n times per year. As n approaches infinity, the effective annual rate approaches e r , where r is the annual interest rate. Understanding such limits helps in modeling and predicting the growth of investments under continuous compounding scenarios, providing a more accurate representation of returns compared to discrete compounding intervals.
