Can someone show me how to evaluate lim x->0+ (1+X)^(1/x) using L'Hopital's rule? I know I need to convert it into proper form, a fraction I think, to solve it. I think it has something to do with...
Can someone show me how to evaluate lim x->0+ (1+X)^(1/x) using L'Hopital's rule?
I know I need to convert it into proper form, a fraction I think, to solve it. I think it has something to do with logarithms but I can't figure it out. Thanks!
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You need to take the logarithm of the limit such that:
`ln lim _(x->0) (1+x)^(1/x) = lim_(x->0) ln ((1+x)^(1/x))`
Using logarithmic identity yields:
`lim_(x->0) ln ((1+x)^(1/x)) = lim_(x->0) (1/x)ln (1+x)`
`lim_(x->0) (1/x)ln (1+x) = lim_(x->0) (ln (1+x))/x = (ln 1)/0 = 0/0`
You may use l'Hospital's theorem such that:
`lim_(x->0) (ln (1+x))/x = lim_(x->0) ((ln (1+x))')/(x') `
`lim_(x->0) ((ln (1+x))')/(x') = lim_(x->0) (1/(1+x))/1`
Substituting 0 for x yields:
`lim_(x->0) ((ln (1+x))')/(x') = 1/(1+0) = 1`
Hence, evaluating the limit logarithm of the limit yields:
`ln lim _(x->0) (1+x)^(1/x) = 1 => lim _(x->0) (1+x)^(1/x) = e^1`
`lim _(x->0) (1+x)^(1/x) = e`
Hence, evaluating the limit, using logarithm and l'Hospital's theorem, yields `lim _(x->0) (1+x)^(1/x) = e` .
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