Apr 5, 2008 #1 rene Messages 1 Reaction score 0 Can somebody solve this problem: the limit of [1 + sin(x)]^(1/x) when x approaches 0 ?
Apr 5, 2008 #2 awvvu Messages 188 Reaction score 1 Take the logarithm of that expression, which will let you use L'Hopital to solve it.
Apr 6, 2008 #3 arildno Science Advisor Homework Helper Gold Member Dearly Missed Messages 10,119 Reaction score 138 The simplest is to rewrite this as: (1+\sin(x))^{\frac{1}{x}}=((1+\sin(x))^{\frac{1}{\sin(x)}})^{\frac{\sin(x)}{x}} The correct limit is quite easy to deduce from this.
The simplest is to rewrite this as: (1+\sin(x))^{\frac{1}{x}}=((1+\sin(x))^{\frac{1}{\sin(x)}})^{\frac{\sin(x)}{x}} The correct limit is quite easy to deduce from this.