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Is the function $g(t)$ bounded or unbounded

$$g(t)= \sqrt {1-\frac 1{t+1}}$$

Should I compute $\lim_{t \to \infty} g(t)$?

And if I do so I get the answer $1$, meaning that the function is bounded(?) but the correct answer according to the textbook is that the function is not bounded.

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  • $\begingroup$ do you mean $$g(t)=\sqrt{1-\frac{1}{t+1}}?$$ $\endgroup$ Sep 2, 2017 at 13:25
  • $\begingroup$ Yes, thank you. $\endgroup$
    – P.ython
    Sep 2, 2017 at 13:26
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    $\begingroup$ Being bounded is a condition on all $x$ in the domain, not just near $\infty$. For instance the function $\frac 1x$ is unbounded even though the limit at $\infty$ is $0$. $\endgroup$
    – lulu
    Sep 2, 2017 at 13:27

3 Answers 3

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The function is unbounded. An intuitive way of seeing this is to consider $t$ to be just below $-1$. Then, $t+1$ will be just below $0$, but we can get arbitrarily close.

To formalize this idea, we take $t = -1 - \epsilon$ for a small $\epsilon > 0$. Then, $g(t) = \sqrt{1 + \frac{1}{\epsilon}}$. Since $\epsilon$ can get as close to $0$ as we want it to be, we can make $\frac{1}{\epsilon}$ as big as we want, and hence we can make $g(t)$ as big as we want.

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note that $$\lim_{t\to \infty}g(t)=1$$ further $$\lim_{t\to -\infty}g(t)=0$$ and $$\lim_{t\to -1^-}g(t)=\infty$$

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$$ g (t)^2=1-\frac {1}{1+t}=\frac {t}{t+1} $$

the domain is $(-\infty,-1)\cup [0,+\infty)$.

$$\lim_{-1^-}g^2 (t)=+\infty $$

$\implies g^2 $ is not bounded

$\implies g $ is unbounded.

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