[catalan]

Hello everyone,

Could someone please comment if my proof below suffices? In my textbook, when was proved using Squeeze Theorem, the inequalities determined were solved explicitly for x. However, my proof below doesn't involve that.

Thank you for your help!

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[mavropnevma]

Perfect. More generally, for any real function , bounded in a neighborhood of zero, by a similar proof .

[debanik2]

Which book is this problem from?

[debanik2]

name the book

[catalan]

Thank you for your response, mavropnevma.

I just want to clarify one question of mine: For squeeze theorem to be applied, the inequality does not have to be strict? Is this because if the inequality is weak, then the limit of the function squeezed (I'll call it ) between the functions ( and ) can be the same as the limits of and ?

For the proof that , the inequalities are strict so I want to understand that strict inequalities may not be compulsory for the Squeeze Theorem to be used.

[mavropnevma]

Of course not. The inequalities may be given (or computed) strict, or not - it makes no difference. The only thing to be careful about is that a strict inequality may become an equality when passing to the limit. The simplest example is , but .

[catalan]

Thank you very much for your response!