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Consider $f(x) = \ln \ x - x$ where $\ln \ x$ refers to base $e$ logarithm. Find limit of $f(x)$ when $x$ goes to infinity.

My try: I know that $f'(x) = \frac{1}{x} - 1$ but what's the next step?

Please Help!

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    Use the fact that $(\log x) /x\to 0$ as $x\to\infty$. – 2017-01-10

4 Answers 4

1

We have $\ln(x) - x$. Factoring out $x$, we get:

$x (\frac{\ln(x)}{x} - 1)$ We know as $x \to \infty$, the first term becomes $\infty$

What about $(\frac{\ln(x)}{x} - 1)?$

If we consider $(\frac{\ln(x)}{x})$, we get an indeterminate form $\frac{\infty}{\infty}$. Applying L'Hopital, we can rewrite $(\frac{\ln(x)}{x})$ in terms of derivative as $(\frac{\frac{1}{x}}{1})$. Evaluating this derivative as $x \to \infty$ gives us $0$.

So we have:

$\infty(0 - 1) = \infty(-1) = -\infty$

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    @K Split X Are you sure that $\infty(0 - 1) = \infty(-1) = -\infty$ is true ? Because we are working with $-\infty$ – 2017-01-10
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    It is true since infinity is simply a really big positive number. Even if you don't think its a number, it's really big and positive. We know $\infty \times \infty = \infty$, and similarly, $1 - \infty = -\infty$. It is one minus a really large positive thing, which makes it negative. So yes, it is true. – 2017-01-10
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Let $f(x)=e^{\ln x-x}$, hence $f(x)=\frac{x}{e^x} \to 0$ for $x \to \infty$.

Thus $ \ln x-x \to - \infty$ for $x \to \infty$.

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    Your answer is good but can you solve it using determining sign of $f'(x)$ ? – 2017-01-10
1

$\ln(x)$ is a function that is strictly increasing on $\mathbb R^+$, but it "flattens out".

$x$ is also strictly increasing, but it keeps a constant positive slope.

This means that $x$ will increase infinitely more than $\ln(x)$.

Thus, $\ln(x)-x \to -\infty$ as $x\to\infty$.

Alternative method using the derivative you found

You know that $f'(x) = \frac1x - 1$, and what we can see from this is that when $x\to\infty$, then $\frac1x \to 0$, and as we subtract 1, the slope will always be negative, and thus the function will tend towards $-\infty$.

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You can use your observation for the derivative. Note that $$ f'(x) = \frac 1x - 1 \to -1, \quad x\to \infty $$ Hence, we can choose an $N$ such that $$ f'(x) < -\frac 12, \qquad x \ge N. $$ This allows us to estimate $$ f(x) = f(N) + \int_N^x f(y)\, dy < f(N) - \frac 12(x-N) \to -\infty, \qquad x \to \infty $$

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    In fact, we may use $N = 2$. Or actually, the way you've used $<$ and $\geq$, any number larger than $2$, but not $2$ itself. – 2017-01-10
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    I dont think hes fimilar with integration – 2017-01-10