1) lim _(xarrow 0)(xe^x)/(e^x)-1

To evaluate the limit \(\lim_{x \rightarrow 0} \frac{xe^x}{e^x - 1}\), we can use L'Hôpital's Rule since the expression is in an indeterminate form \(\frac{0}{0}\).<br /><br />First, let's apply L'Hôpital's Rule. L'Hôpital's Rule states that if \(\lim_{x \rightarrow c} \frac{f(x)}{g(x)}\) results in an indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then:<br /><br />\[<br />\lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \lim_{x \rightarrow c} \frac{f'(x)}{g'(x)}<br />\]<br /><br />provided the limit on the right-hand side exists. Here, \(f(x) = xe^x\) and \(g(x) = e^x - 1\).<br /><br />First, we need to find the derivatives of \(f(x)\) and \(g(x)\):<br /><br />\[<br />f(x) = xe^x \implies f'(x) = e^x + xe^x = e^x(1 + x)<br />\]<br /><br />\[<br />g(x) = e^x - 1 \implies g'(x) = e^x<br />\]<br /><br />Now, applying L'Hôpital's Rule:<br /><br />\[<br />\lim_{x \rightarrow 0} \frac{xe^x}{e^x - 1} = \lim_{x \rightarrow 0} \frac{e^x(1 + x)}{e^x}<br />\]<br /><br />We can cancel out \(e^x\) in the numerator and the denominator:<br /><br />\[<br />\lim_{x \rightarrow 0} (1 + x)<br />\]<br /><br />Now, substitute \(x = 0\):<br /><br />\[<br />1 + 0 = 1<br />\]<br /><br />Therefore, the limit is:<br /><br />\[<br />\boxed{1}<br />\]