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`lim_(x->0+)(1+5/x)^x` 0+ is the right hand limit Please give your answer to 4...

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kitkitbb | eNotes Newbie

Posted October 27, 2013 at 3:06 AM via web

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`lim_(x->0+)(1+5/x)^x`

0+ is the right hand limit

Please give your answer to 4 decimal place.

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1 Answer | Add Yours

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mvcdc | Student , Undergraduate | (Level 1) Associate Educator

Posted October 27, 2013 at 4:34 AM (Answer #1)

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The function is continuous at 0 so the left hand limit is just the same as the right hand limit, and the limit exists at 0.

`lim_(x -> 0) (1+5/x)^x`

This is an indeterminate for of type `infty^0` . To resolve this, we perform the following transformation:

`lim_(x->0)(1+5/x)^x = e^(lim_(x->0)log(1+5/x)^x) = e^(lim_(x->0)xlog(1+5/x))` , using also the property of logarithms. 

The resulting expression is still an indeterminate, this time, of type `0*infty` . We perform a substitution. Let `a = 1/x` . We now have:

`e^(lim_(x->0)xlog(1+5/x)) = e^(lim_(a -> +infty) (log(1+5a))/(a))`

This is still an indeterminate form, `infty/infty` but is easier to resolve. We simply have to use L'Hospital's Rule:

`lim_(a->+infty) (log(1+5a))/a = lim_(a->+infty) ((d(log(1+5a)))/(da))/((d(a))/(da))`

`lim_(a->+infty) (5/(1+5a))`

Note that the limit of a quotient is just the quotient of the limits, and the limit of the constant is same as the constant. Also, remember to bring back the whole expression (base e):

`e^(5/(lim_(a->+infty)(1+5a)))` 

`lim_(a->+infty)(1+5a)` is infinity, so the entire exponent approaches 0. Therefore:

`lim_(x->0)(1+5/x)^x = e^0 = 1`

[Note that log here is the natural logarithm, ln]

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