is this function contineous

**pravinjaggu** · August 20th 2009, 08:24 PM

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??

**Chris L T521** · August 20th 2009, 08:27 PM

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.

**VonNemo19** · August 20th 2009, 08:27 PM

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??