It was already mentioned that infinity is not a number. Therefore in arithmetic and algebra the answer would be: UNDEFINED.
In calculus, in a section involving " limits " the idea of " approaching " is introduces. For example:
lim 2x = ?...... The question is : " What is the limit of f (x) = 2x as x approaches 5 ?.
x -> 5
In plane language: " When x approaches the value 5, what value does 2x approaching ( if any ) ?
If f(x) = 2x approaches a unique number as x approaches 5, that number is the limit. In this case ( a very basic example ) when x takes the values :
......................... x ........... f(x)
......................... 4 ........... 8
........................ 4.5 .......... 9
........................ 4.75 .........9.5
.........................4.9 .......... 9.8
........................ 4.99......... 9.99
........................ 4.99999.... 9.99998
........................ 5.000001...10.000002
........................ 5.00001.... 10.00002
........................ 5.0001 ..... 10.0002
We x approaches 5 from both sides below and above, 2x looks more and more like 10. Note that x is never equal to 5. X can approach 5 for ever.
We then say : When x approaches 5; the function f (x) = 2x approaches 10
Now your question:
lim 1/ x = ?..... What is the limit of f (x) = 1/x as x approaches oo ( Gets larger and larger forever )
.. x-> oo
We know that if the numerator of a fraction is constant and the denominator of the same fraction is getting larger, the value of the fraction is getting smaller:
For example: 1/ 10 > 1/100 > 1/ 1000, etc.
Therefore: if the numerator of a fraction is constant and equal to 1 and the denominator is getting larger and larger for ever, the value of the fraction is getting is getting smaller and smaller forever. That is is" getting closer and closer to cero (looking more and more like cero) " It is never equal to cero.
Note : Of course calculus has its own tools, that make the proccess shorter.