Which one of the following functions is strictly bounded?
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$\frac{1}{x^2}$
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ex
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x2
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$e^{-x^2}$
D
Correct answer
Explanation
$\text{We have} \hspace{1cm} lim_{x \rightarrow 0} \frac{1}{x^2} = \infty
\\
\hspace{2.7cm} lim_{x \rightarrow \infty} x^2 = \infty
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\hspace{2.7cm} lim_{x \rightarrow \infty} e^{-x} = \infty
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\hspace{2.7cm} lim_{x \rightarrow \infty} e^{-x^2} = 0
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\hspace{2.7cm} lim_{x \rightarrow 0} e^{-x^2} = 1 \hspace{1cm} \text{Thus $e^{-x^2}$ is strictly bounded.}$