I already know many resolutions for this problem (which $y=\frac{1}{2}$ was the answer), but I was not able to find where I did a mistake in my original one: \begin{align} y &=\lim_{x \to \infty} \left( x - x^2 \ln{\frac{(x+1)}{x}} \right)\\\\ &=\lim_{x \to \infty} x \ \cdot \lim_{x \to\infty} \left( 1 - x \ln{\left(1+\frac{1}{x}\right)} \right) \end{align} Then,I tried to use l'hopital on the right limit(after some manipulations, such as spliting the limit in two (for each side of the subtraction) and "tossing" that x to the denominator in its inverse form so that I had $\frac{0}{0}$), but I ended up with \begin{align} \lim_{x \to\infty} \frac{1}{x+1} \end{align} Which would, in the end (multiplying again with the left limit I left in the beginning): \begin{align} \lim_{x \to\infty} \frac{x}{x+1} = 1 \end{align}
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5$\begingroup$ Your second line is invalid. $\lim_{x \to a} (f(x) g(x)) = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)$ is only true if both factors on RHS are defined. You end up with $\infty \cdot 0$, which doesn't work. $\endgroup$– dgeyfmanCommented Aug 11, 2023 at 19:32
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1$\begingroup$ They didn't really end up using $\lim(fg)=\lim(f)\lim(g)$. So, not really the problem. The problem is in the part not mentioned, in obtaining the asymptotic $\frac{1}{x+1}$. $\endgroup$– NDBCommented Aug 11, 2023 at 19:41
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1$\begingroup$ @excitedGoose The factor $1-x\log(1+1/x)$ has Taylor expansion (at $\infty$) of the form $\frac{1}{2x}+o(1/x)$, not $\frac{1}{x+1}$. So, when you multiply by $x$, you get $\frac{1}{2}+o(1)\to\frac{1}{2}$ $\endgroup$– NDBCommented Aug 11, 2023 at 19:46
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You fell into a trap for young players. The limit porpriety $$ \lim_{x\to a} f(x)g(x) = \lim_{x\to a} f(x) \cdot \lim_{x\to a} g(x) $$ is only true if the individual limits for $f(x)$ and $g(x)$ exist and are finite. Consider, as a trivial example, the limit $$ \lim_{x\to 0} \frac{x}{x} $$ which can be shown to be $1$ with no effort. However, if you tried to evaluate it as $$ \lim_{x\to 0} \frac{x}{x} = \lim_{x\to 0} x \cdot \lim_{x\to 0} \frac{1}{x} $$ you could be tempted to say that it doesn't exists because $$ \lim_{x\to 0} \frac{1}{x} $$ doesn't.
