# Thread: is this function contineous

1. ## is this function contineous

i have a question.
is a function limit x tending to 1 f(x)= (x^2-1)/(x-1)
is contineous at every point or it is discontineous at x=1?

also the simplified form of above function i.e (x+1) has the same contineuity as the above function??

2. Originally Posted by pravinjaggu
i have a question.
is a function limit x tending to 1 f(x)= (x^2-1)/(x-1)
is contineous at every point or it is discontineous at x=1?

also the simplified form of above function i.e (x+1) has the same contineuity as the above function??
$\lim_{x\to1}\frac{x^2-1}{x-1}=\lim_{x\to1}\frac{\left(x+1\right)\left(x-1\right)}{x-1}=\lim_{x\to1}x+1=2$.

However, $f\left(x\right)=\frac{x^2-1}{x-1}$ is discontinuous at $x=1$. There is a hole in the function at that value since $x-1$ is a term common to the numerator and denominator of the function.

3. Originally Posted by pravinjaggu
i have a question.
is a function limit x tending to 1 f(x)= (x^2-1)/(x-1)
is contineous at every point or it is discontineous at x=1?

also the simplified form of above function i.e (x+1) has the same contineuity as the above function??
$f(x)$ is discontinuos at $x=1$ because it is not DEFINED there.

This is a REMOVABLE discontinuity because

$f(x)=x+1\forall{x\neq1}$

4. Originally Posted by VonNemo19
$f(x)$ is discontinuos at $x=1$ because it is not DEFINED there.

This is a REMOVABLE discontinuity because

$f(x)=x+1\forall{x\neq1}$
Hello there is the graphe