Limit of [1 + sin(x)]^(1/x) when x approaches 0

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In summary, the limit of [1 + sin(x)]^(1/x) when x approaches 0 is equal to 1. To find the limit, one can use L'Hôpital's rule or the power rule for limits. This limit is not equal to 0, but rather represents the value that the function approaches as x approaches 0, which is 1.
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rene
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Can somebody solve this problem: the limit of [1 + sin(x)]^(1/x) when x approaches 0 ?
 
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  • #2
Take the logarithm of that expression, which will let you use L'Hopital to solve it.
 
  • #3
The simplest is to rewrite this as:
[tex](1+\sin(x))^{\frac{1}{x}}=((1+\sin(x))^{\frac{1}{\sin(x)}})^{\frac{\sin(x)}{x}}[/tex]
The correct limit is quite easy to deduce from this.
 

1. What is the limit of [1 + sin(x)]^(1/x) when x approaches 0?

The limit of [1 + sin(x)]^(1/x) when x approaches 0 is equal to 1.

2. How do you find the limit of [1 + sin(x)]^(1/x) when x approaches 0?

To find the limit of [1 + sin(x)]^(1/x) when x approaches 0, you can use the L'Hôpital's rule or the power rule for limits.

3. Is the limit of [1 + sin(x)]^(1/x) when x approaches 0 equal to 0?

No, the limit of [1 + sin(x)]^(1/x) when x approaches 0 is not equal to 0. It is equal to 1.

4. Does the limit of [1 + sin(x)]^(1/x) when x approaches 0 exist?

Yes, the limit of [1 + sin(x)]^(1/x) when x approaches 0 does exist and it is equal to 1.

5. What does the value of the limit of [1 + sin(x)]^(1/x) when x approaches 0 represent?

The value of the limit of [1 + sin(x)]^(1/x) when x approaches 0 represents the value that the function approaches as x gets closer and closer to 0. In this case, it represents the value of 1.

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