Lt [x.e^x-log(1+x)]/x^2 as x tends to 0.

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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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Evaluate: lim(x →0) (xe^x - log(1 + x))/x2 Evaluate: lim(x →0) (xe^x - log(1 + x))/x2
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
Lt x tends to 0 [xe^x-log(1+x)]/x^2 Evaluate Lt x tends to 0 [xe^x-log(1+x)]/x^2 Evaluate
Evaluate: lim(x→0) (xe^x –log(1-x))/x^2 using l hospital rule Evaluate: lim(x→0) (xe^x –log(1-x))/x^2 using l hospital rule
example 9 show that limit x tend to 0 (e^x+log(1-x/e))/(tanx-x) = -1/2 use l hospital rule example 9 show that limit x tend to 0 (e^x+log(1-x/e))/(tanx-x) = -1/2 use l hospital rule