Given to solve ,

`lim_(x->oo) e^(x/2)/x`

As `x` thends to ` oo` we get `e^(x/2)/x = oo/oo`

L'Hopital's Rule says if

`lim_(x->a) f(x)/g(x) = 0/0` or `(+-oo)/(+-oo)` then the limit is: `lim_(x->a) f'(x)/g'(x)`

so , now evaluating

`lim_(x->oo) e^(x/2)/x`

upon using the L'Hopital's Rule we get

=`lim_(x->oo) ((e^(x/2))')/((x)')`

=`lim_(x->oo) ((e^(x/2))(1/2))/(1)`

=>`lim_(x->oo)...

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Given to solve ,

`lim_(x->oo) e^(x/2)/x`

As `x` thends to ` oo` we get `e^(x/2)/x = oo/oo`

L'Hopital's Rule says if

`lim_(x->a) f(x)/g(x) = 0/0` or `(+-oo)/(+-oo)` then the limit is: `lim_(x->a) f'(x)/g'(x)`

so , now evaluating

`lim_(x->oo) e^(x/2)/x`

upon using the L'Hopital's Rule we get

=`lim_(x->oo) ((e^(x/2))')/((x)')`

=`lim_(x->oo) ((e^(x/2))(1/2))/(1)`

=>`lim_(x->oo) (e^(x/2))/2`

now on` x-> oo` we get `e^(x/2) -> oo`

so,

`lim_(x->oo) (e^(x/2))/2 = oo`