Contradiction of computation of this:$\lim _{x\to 0^+} (\frac{({2}^{(e^{x^\frac32}+1)}-4)^2}{\sin^6{(\sqrt{x}})})$ with wolfram alpha?

Question

I have claculated :$\lim _{x\to 0^+} (\frac{({2}^{(e^{x^\frac32}+1)}-4)^2}{\sin^6{(\sqrt{x}})})$ I have got $( 4\log 2))^2$ as shown below ,but wolfram alpha assumed it equal $\infty$ , then my question here where is the problem in my steps of evaluation ?

Answer 1

Your mistake is a bracket misplaced in what you wrote in wolfram alpha