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Answer :
Final answer:
The limit of the function [tex](1/x)^(tanx)[/tex] as x approaches 0 is e. This is achieved by using L'Hopital's Rule after getting an indeterminate form and rewriting the function in a compatible form by taking the natural logarithm.
Explanation:
The equation given is a calculus problem related to limits and more specifically to L'Hopital's rule. The given equation is [tex]lim (1/x)^(tanx)[/tex] as x approaches 0. Attempting to plug 0 into x won't work in this equation as it creates an indeterminate form, so we use L'Hopital's Rule, which states if the limit of a function as x approaches a certain value results in an indeterminate form of 0/0 or ∞/∞, we can take the derivative of the top and bottom separately to come up with a new limit that should give a definitive answer. But first, we need to rewrite the given function into a form where we can use L'Hopital's Rule.
We do this by taking the natural logarithm of both sides. Since our function is in an exponent, taking the natural logarithm allows us to pull the exponent in front. So we get: ln(y) = tanx * ln(1/x). Instead of finding the limit of the original function, we will find the limit of this one and then exponentiate the result at the end. The limit of this new function as x approaches 0 is of the indeterminate form 0/0. Now we can use L'Hopital's Rule: take the derivative of the top and the bottom function. For the derivative of the top function (tanx), we use the identity sec²x, and for the bottom function (ln(1/x)), we get -1/x. The limit of this function as x approaches 0 is 1.
The concluding answer is obtained by undoing the logarithm we introduced via exponentiation, bringing the result back to our original function. Thus, e⁽ˡⁱᵐ[ˣ→⁰]⁽(tanx * ln(1/x))⁾ = e⁽¹⁾ = e. Therefore, the limit of the function [tex](1/x)^(tanx)[/tex] as x approaches 0 is e.
Learn more about L'Hopital's Rule here:
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