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??
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??
$\displaystyle \lim_{x\to1}\frac{x^2-1}{x-1}=\lim_{x\to1}\frac{\left(x+1\right)\left(x-1\right)}{x-1}=\lim_{x\to1}x+1=2$.
However, $\displaystyle f\left(x\right)=\frac{x^2-1}{x-1}$ is discontinuous at $\displaystyle x=1$. There is a hole in the function at that value since $\displaystyle x-1$ is a term common to the numerator and denominator of the function.