Infinity Question

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?

#Calculus

Easy Math Editor

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Comments

The $2^{\text{nd}}$ 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

Hamza A - 5 years ago

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 - 5 years ago

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 - 5 years ago

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 - 5 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$