`lim_(x->0^+) (4x + 1)^cot(x)` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t...

`lim_(x->0^+) (4x + 1)^cot(x)` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

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Textbook Question

Chapter 4, 4.4 - Problem 63 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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nees101 | Student, Graduate | (Level 2) Adjunct Educator

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Given the function `lim_{x->0^+}(4x+1)^cot(x)`

We have to find the limit.

The given function is an intermediate form of type `1^infty` which can be written as:

i.e `lim_{x->0^+}f(x)^g(x)=e^(lim_{x->0^+}g(x)[f(x)-1])`

`e^(lim_{x->0^+}g(x)[f(x)-1])=e^(lim_{x->0^+}cot(x)[4x+1-1])=e^(lim_{x->0^+}cot(x)4x)`

                                                             `=e^(lim_{x->0^+}(4cos(x).x)/sin(x))`

                                                              `=e^(lim_{x->0^+}(4cos(x))/(sin(x)/x))`

We know that `lim_{x->0}sin(x)/x=1`

Therefore we get,

`e^(lim_{x->0^+}4cos(x))=4`

Therefore the limit is e^4.

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