rene Messages 1 Reaction score 0 Thread starter Apr 5, 2008 #1 Can somebody solve this problem: the limit of [1 + sin(x)]^(1/x) when x approaches 0 ?
awvvu Messages 188 Reaction score 1 Apr 5, 2008 #2 Take the logarithm of that expression, which will let you use L'Hopital to solve it.
arildno Science Advisor Homework Helper Gold Member Dearly Missed Messages 10,165 Reaction score 138 Apr 6, 2008 #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.
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.