I/ lim _(x arrow infty)(e^x-x) [ 2 / lim _(x arrow-infty) (e^x)/(e^x)-x 3 / lim _(x arrow+infty) e^-5 x+3 ]

1/ \( \lim _{x \rightarrow \infty}\left(e^{x}-x\right) \)To find the limit of as approaches infinity, we can use L'Hôpital's Rule. L'Hôpital's Rule states that if the limit of a function is in the form of or , then the limit of the derivative of the numerator divided by the derivative of the denominator is equal to the original limit.In this case, as approaches infinity, both and will increase without bound. Therefore, the limit is in the form of .To apply L'Hôpital's Rule, we need to find the derivatives of the numerator and denominator.The derivative of is , and the derivative of is 1.So, applying L'Hôpital's Rule, we have: Therefore, the limit of as approaches infinity is .2/ To find the limit of as approaches negative infinity, we can again use L'Hôpital's Rule.As approaches negative infinity, approaches 0, and approaches negative infinity. Therefore, the limit is in the form of .To apply L'Hôpital's Rule, we need to find the derivatives of the numerator and denominator.The derivative of is , and the derivative of is .So, applying L'Hôpital's Rule, we have: Therefore, the limit of as approaches negative infinity is 0.3/ To find the limit of as approaches positive infinity, we can use the fact that the exponential function approaches 0 as its exponent approaches negative infinity.As approaches positive infinity, approaches negative infinity. Therefore, the limit is in the form of .Since the exponential function approaches 0 as its exponent approaches negative infinity, we have: Therefore, the limit of as approaches positive infinity is 0.