L'Hôpital's Rule states that for functions f(x) and g(x) that are differentiable on an open interval I, except possibly at a point c contained in I, if the limit as x approaches c of f(x) and g(x) is 0 or ±∞, then the limit as x approaches c of [f(x) / g(x)] is equal to the limit as x approaches c of [f'(x) / g'(x)], provided the latter limit exists.

Now, let's evaluate the limit lim x - 0 (1 + sin 4x) ^ (cot x).

First, we rewrite the expression in the form of 0/0 or ∞/∞ to apply L'Hôpital's Rule. We can rewrite the expression as e^(cot(x) * ln(1 + sin(4x))), which simplifies to e^(ln((1 + sin(4x))^(cot(x)))).

This is equivalent to the original expression, but now we can apply L'Hôpital's Rule to the exponent, cot(x) * ln(1 + sin(4x)), because as x approaches 0, both cot(x) and ln(1 + sin(4x)) approach ∞.

Taking the derivative of the numerator, ln(1 + sin(4x)), gives (4cos(4x)) / (1 + sin(4x)).

Taking the derivative of the denominator, cot(x), gives -csc^2(x).

So, the limit as x approaches 0 of the exponent is the limit as x approaches 0 of [(4cos(4x)) / (1 + sin(4x))] / -csc^2(x).

Applying L'Hôpital's Rule again, we find that this limit is equal to the limit as x approaches 0 of [(-16sin(4x)) / (1 + sin(4x))^2] / 2cot(x)csc^2(x).

This limit is 0, so the limit as x approaches 0 of the original expression, (1 + sin 4x) ^ (cot x), is e^0 = 1.

Question

L'Hôpital's Rule states that for functions f(x) and g(x) that are differentiable on an open interval I, except possibly at a point c contained in I, if the limit as x approaches c of f(x) and g(x) is 0 or ±∞, then the limit as x approaches c of [f(x) / g(x)] is equal to the limit as x approaches c of [f'(x) / g'(x)], provided the latter limit exists.

Now, let's evaluate the limit lim x -> 0 (1 + sin 4x) ^ (cot x).

First, we rewrite the expression in the form of 0/0 or ∞/∞ to apply L'Hôpital's Rule. We can rewrite the expression as e^(cot(x) * ln(1 + sin(4x))), which simplifies to e^(ln((1 + sin(4x))^(cot(x)))).

This is equivalent to the original expression, but now we can apply L'Hôpital's Rule to the exponent, cot(x) * ln(1 + sin(4x)), because as x approaches 0, both cot(x) and ln(1 + sin(4x)) approach ∞.

Taking the derivative of the numerator, ln(1 + sin(4x)), gives (4cos(4x)) / (1 + sin(4x)).

Taking the derivative of the denominator, cot(x), gives -csc^2(x).

So, the limit as x approaches 0 of the exponent is the limit as x approaches 0 of [(4cos(4x)) / (1 + sin(4x))] / -csc^2(x).

Applying L'Hôpital's Rule again, we find that this limit is equal to the limit as x approaches 0 of [(-16sin(4x)) / (1 + sin(4x))^2] / 2cot(x)csc^2(x).

This limit is 0, so the limit as x approaches 0 of the original expression, (1 + sin 4x) ^ (cot x), is e^0 = 1.

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