Question #6c442

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Answer 1 · 2017-02-21T12:49:22

$lim_(x->oo) (1+1/x)^x = e$

Write the function as:

$(1+1/x)^x = (e^ln(1+1/x))^x = e^(xln(1+1/x))$

We evaluate now:

$lim_(x->oo) xln(1+1/x)$

This is in the indeterminate form $0*oo$ , but we can transform it to the form $0/0$ and solve it using l'Hospital's rule:

$lim_(x->oo) xln(1+1/x) = lim_(x->oo) ln(1+1/x)/(1/x)$

$lim_(x->oo) xln(1+1/x) = lim_(x->oo) (d/dx ln(1+1/x))/(d/dx (1/x))$

$lim_(x->oo) xln(1+1/x) = lim_(x->oo) (1/(1+1/x)*(-1/x^2))/(-1/x^2)$

$lim_(x->oo) xln(1+1/x) = lim_(x->oo) 1/(1+1/x) = 1$

Now, as $e^x$ is a continuous function we have:

$lim_(x->oo) e^(xln(1+1/x)) = e^((lim_(x->oo) xln(1+1/x))) = e^1 = e$

So, in conclusion:

$lim_(x->oo) (1+1/x)^x = e$