# Use L'Hospital's rule to solve: `lim_ (x-> +oo) ` `e^(-x) - 1/2x`

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### 1 Answer

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

Let's express the given function in rational form to be able to apply L'Hospital's rule.

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

From here, let's apply the L'Hospital's rule. So, take the derivative of the numerator and the denominator.

`=lim_(x->+oo)` `(0-(x*e^x+e^x*1))/(2e^x*1)` `= lim_(x->+oo)` `-(xe^x+e^x)/(2e^x)`

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

Then, apply the rule `lim_(x->oo) c/x^r =0` , where c is a constant and r is an exponent whose value is greater than zero.To do so, divide all the terms by x.

`= lim_(x->+oo)` `- (x/x+1/x)/(2/x)` `= lim_(x->+oo) -(1+1/x)/(2/x)` `= -(1+0)/0 = -1/0`

Note that if we divide the number by zero, the quotient is infinity.

`= -1*oo`

Then, follow the rules for multiplication of unlike signs. So, if we multiply -1 with a positive large number, the product would have a large value with a negative sign. So we have,

`=-oo`

**Hence, `lim_(x->+oo)``e^(-x)-1/2x = -oo` . **

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