Please show all work in order to receive 5 stars. Thanks!! limx rightarrow 0(1+x

Question

Please show all work in order to receive 5 stars.

Thanks!!

Explanation / Answer

Take the limit: lim_(x-%u026C) (1+0.5 x)^(4/x) Indeterminate form of type 1^infinity. Transform using lim_(x->0) (1+0.5 x)^(4/x) = e^(lim_(x->0) (4 log(1+0.5 x))/x): = e^(lim_(x->0) (4 log(1+0.5 x))/x) Factor out constants: = e^(4 (lim_(x->0) (log(1+0.5 x))/x)) Indeterminate form of type 0/0. Applying L'Hospital's rule we have, lim_(x->0) (log(1+0.5 x))/x = lim_(x->0) (( dlog(1+0.5 x))/( dx))/(( dx)/( dx)): = e^(4 (lim_(x->0) 0.5/(1+0.5 x))) Factor out constants: = e^(2. (lim_(x->0) 1/(1+0.5 x))) The limit of a quotient is the quotient of the limits: The limit of a constant is the constant: = e^(2./(lim_(x->0) (1+0.5 x))) The limit of 1+0.5 x as x approaches 0 is 1.: Answer: | | = 7.38906

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