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Find the value of alpha so that ("lim")(...

Find the value of `alpha` so that `("lim")_(xvec0)1/(x^2)(e^(alphax)-e^x-x)=3/2`

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The correct Answer is:
2
Since the numerator tends to 0 as `xrarr0` so `lim_(xrarr0)((e^(ax)-e^(x)-x)/(x^2))=1/2lim_(xrarr0)((ax^(ax)-e^(x)-1))/(x)`
For the last limit to exist we must have `lim_(xrarr0)(ae^(ax)-e^(x)-1)=0 " ":. " "alpha-1-1=0implies alpha =2`
For `alpha = 2` , the last limit is equal to `1/2lim_(xrarr0)((2e^(2x)-e^(x)-1))/(x)=1/2 lim_(xrarr0)(4e^(2x)-e^x)=3/2`
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