Why does \(\displaystyle \lim _{ x\rightarrow \infty }{ \frac { 8 }{ x } } =0\) but \(\displaystyle \lim _{ x\rightarrow \infty }{ x\cdot 0 } =8\) is not true? What kind of number is zero?

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TopNewestThe \(2^{\text{nd}}\) limit isn't a valid one,the function you're taking the limit of is \(0\) for all \(x\) (anything multiplied by \(0\) will give \(0\)),but the first one is valid since the function \(\frac{8}{x}\) approaches \(0\) as \(x\rightarrow\infty\).I think you multiplied both sides by \(x\) then switched the limit to the \(\text{RHS}\),this is not valid as you can't multiply by the variable in a limit equation as you did,multiplying by a constant is valid though.

and to answer your second question,it's a whole number – Hummus A · 6 months, 1 week ago

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One of the problems if we went ahead and accepted the second limit to be true is that we can replace 8 with another number and still get an equivalent answer.

So, the second limit would be simultaneously equal to two numbers - which creates a contradiction, which is not acceptable in mathematics. – Star Light · 6 months ago

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I think 0 is not a number. It shows there is nothing.Just like infinite which shows there is everything. But other numbers show something. – Akash Shukla · 6 months, 1 week ago

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I think you can't do this because the limit is written

\(\lim _{ x \rightarrow x_0} {f(x)}=l \)

and so 8 and x are strictly linked (they are f(x)) in your equation and you can't do what you have done. – Matteo Monzali · 6 months, 1 week ago

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