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# Evaluate$$\displaystyle \lim_{x\to \infty} \left[ x - x^2 \ln \left( \frac{1 + x}{x} \right) \right]$$.

## The limit, $L=\lim _{x \rightarrow \infty}\left[x-x^{2} \ln \left(\frac{1+x}{x}\right)\right]=\lim _{x \rightarrow \infty}\left[x-x^{2} \ln \left(\frac{1}{x}+1\right)\right] .$ Let $t=1 / x,$ so as $x \rightarrow \infty, t \rightarrow 0^{+}$$L=\lim _{t \rightarrow 0^{+}}\left[\frac{1}{t}-\frac{1}{t^{2}} \ln (t+1)\right]=\lim _{t \rightarrow 0^{+}} \frac{t-\ln (t+1)}{t^{2}} \stackrel{\mathrm{n}}{=} \lim _{t \rightarrow 0^{+}} \frac{1-\frac{1}{t+1}}{2 t}=\lim _{t \rightarrow 0^{+}} \frac{t /(t+1)}{2 t}=\lim _{t \rightarrow 0^{+}} \frac{1}{2(t+1)}=\frac{1}{2}$Note: Starting the solution by factoring out $x$ or $x^{2}$ Ieads to a more complicated solution.

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this problem, we're gonna be taking the limit as X goes to infinity. But we're going to want to rewrite the given limit and apply a substitution. We're gonna let you equals one over X. So using all of this um substitution, we end up getting it into the form of the limit as X goes to infinity of x minus x squared, I'm the natural law of one over X plus one. Then based on this we end up getting when we plug in U. Equals zero, we get 0/0. So we're gonna have to use local tiles. Rule a substituting you where we can we get one minus one over you plus one and then that's all gonna be over to you. Then Roy plug in zero here. And simplify further. We get um one over two times u plus one, plugging in zero. We get 1/2. So one half is going to be the limit for this function.

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