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If $f(x)$ is strictly increasing, is $\lim_{x\to\infty} f(x) = \infty$ ? Also is $\lim_{x\to -\infty} = 0?$ I think the answer is yes. A good example is $e^{x}$. I don't know how to show this claim though. Is it true?

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    $\begingroup$ No, if the function is bounded the limit is finite. Consider $f(x) =-1/x$ as an example. $\endgroup$ – Paramanand Singh Nov 12 '18 at 6:22
  • $\begingroup$ Then how can I show $\lim_{x\to\infty} e^{x} = \infty$ ? $\endgroup$ – Dillain Smith Nov 12 '18 at 6:25
  • $\begingroup$ @DillainSmith You can prove it directly from the definition of $\lim_{x\to\infty}f(x)=\infty$. $\endgroup$ – Arthur Nov 12 '18 at 6:26
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Use the following theorem:

Theorem: Let $f:[a, \infty) \to\mathbb{R} $ be a function. If $f$ is increasing on $[a, \infty) $ then either $\lim_{x\to\infty} f(x) $ exists or $f(x) \to\infty$ as $x\to\infty $.

The proof can be given by analyzing the range $A=f([a, \infty)) $ of $f$. If $A$ is bounded above (meaning every member of $A$ is less than some fixed number) then $L=\sup A$ exists and one can prove that $f(x) \to L$ as $x\to\infty$ (prove this!).

On the other hand if $A$ is unbounded above ie for any given $K>0$ there is a value $x_0\geq a$ such that $f(x_0)>K$ then by increasing nature of $f$ we have $f(x) \geq f(x_0)>K$ for all $x>x_0$. Thus $f(x) \to \infty $ as $x\to\infty $.

For $f(x) =e^x$ the situation is very simple as it satisfies the inequality $e^x\geq x+1>x$ for all $x$. Hence given any $K>0$ there is an $x_0$ namely $x_0=K$ such that $$f(x)=e^x>x>x_0=K$$ for all $x>x_0$. And thus $f(x) \to\infty $ as $x\to\infty $.

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Hint: What about the $\tanh$ function?

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You know what is $$\lim\limits_{x\to-\infty}e^x?$$What about $$\lim\limits_{x\to\infty}-e^{-x}?$$ And what about $$\lim\limits_{x\to-\infty}-e^{-x}?$$

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What about for $x>0$

$$f(x)=1-\frac1x\implies f’(x)=\frac1{x^2}>0$$

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