$\frac{x}{x}$ is always 1?

Even when x = 0?

I'm learning how to get the asymptote of a rational function and I was curious about the function: $\newline \frac{x-a}{(x-a)(x-b)}$

It only has one vertical asymptote. I understand that the expression can be simplified, but when «x = a» the expression (without simplifying) it remains: $\newline \frac{0}{0(a-b)} = \frac{0}{0}$

Which is not defined.
But with the simplification yes, it is defined.

I would like to know who to believe. I would like to know where I am wrong.

#Algebra

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Comments

Whenever you simplify the same term from the top and bottom, you write the condition that this term should not be equal to 0. Hence, when you simplify $\frac{\cancel{(x-a)}}{\cancel{(x-a)}(x-b)}$ , you also write $x-a \ne 0 \implies x \ne a$ .

So, this rational function is never defined when $x=a$ , so it will be undefined at that point and as pointed by Krishna, it will have a hole. Also, this function will be discontinuous

Mahdi Raza - 5 months, 3 weeks ago

0/0 is never 1. I understand your logic, but it's best to think of such a definition as nonsensical. So x/x is not always 1. So when you represent it on a graph, it is a function with a little hole at x = 0 because x cannot be equal to zero.

Krishna Karthik - 5 months, 3 weeks ago

thanks, i understood perfectly

Emmanuel Jesús - 5 months, 3 weeks ago

No problem mate. Have a good day.

Krishna Karthik - 5 months, 2 weeks 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}`	$\boxed{123}$

xx \frac{x}{x} xx​ is always 1?

Comments

$\frac{x}{x}$ is always 1?