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/ \( \lim _{x \rightarrow \infty}\left(e^{x}-x\right) \)<br /><br />To find the limit of \( e^x - x \) as \( x \) 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 \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then the limit of the derivative of the numerator divided by the derivative of the denominator is equal to the original limit.<br /><br />In this case, as \( x \) approaches infinity, both \( e^x \) and \( x \) will increase without bound. Therefore, the limit is in the form of \( \frac{\infty}{\infty} \).<br /><br />To apply L'Hôpital's Rule, we need to find the derivatives of the numerator and denominator.<br /><br />The derivative of \( e^x \) is \( e^x \), and the derivative of \( x \) is 1.<br /><br />So, applying L'Hôpital's Rule, we have:<br /><br />\[ \lim _{x \rightarrow \infty}\left(e^{x}-x\right) = \lim _{x \rightarrow \infty}\left(\frac{e^x}{1}\right) = \infty \]<br /><br />Therefore, the limit of \( e^x - x \) as \( x \) approaches infinity is \( \infty \).<br /><br />2/ \( \lim _{x \rightarrow-\infty} \frac{e^{x}}{e^{x}-x} \)<br /><br />To find the limit of \( \frac{e^x}{e^x - x} \) as \( x \) approaches negative infinity, we can again use L'Hôpital's Rule.<br /><br />As \( x \) approaches negative infinity, \( e^x \) approaches 0, and \( x \) approaches negative infinity. Therefore, the limit is in the form of \( \frac{0}{-\infty} \).<br /><br />To apply L'Hôpital's Rule, we need to find the derivatives of the numerator and denominator.<br /><br />The derivative of \( e^x \) is \( e^x \), and the derivative of \( e^x - x \) is \( e^x - 1 \).<br /><br />So, applying L'Hôpital's Rule, we have:<br /><br />\[ \lim _{x \rightarrow-\infty} \frac{e^{x}}{e^{x}-x} = \lim _{x \rightarrow-\infty} \frac{e^x}{e^x - 1} = 0 \]<br /><br />Therefore, the limit of \( \frac{e^x}{e^x - x} \) as \( x \) approaches negative infinity is 0.<br /><br />3/ \( \lim _{x \rightarrow+\infty} e^{-5 x+3} \)<br /><br />To find the limit of \( e^{-5x + 3} \) as \( x \) approaches positive infinity, we can use the fact that the exponential function approaches 0 as its exponent approaches negative infinity.<br /><br />As \( x \) approaches positive infinity, \( -5x + 3 \) approaches negative infinity. Therefore, the limit is in the form of \( e^{-\infty} \).<br /><br />Since the exponential function approaches 0 as its exponent approaches negative infinity, we have:<br /><br />\[ \lim _{x \rightarrow+\infty} e^{-5 x+3} = 0 \]<br /><br />Therefore, the limit of \( e^{-5x + 3} \) as \( x \) approaches positive infinity is 0.