Lt x tends to 0 [xe^x-log(1+x)]/x^2 Evaluate

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Using L hospital rule lim x tends to zero x.e^x-log(1+x)/x^2 Using L hospital rule lim x tends to zero x.e^x-log(1+x)/x^2
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limit of (xe^x - log(1 + x))/x^2 as x approaches 0 without application of L’Hospital’s Rule limit of (xe^x - log(1 + x))/x^2 as x approaches 0 without application of L’Hospital’s Rule