Where is $\frac{x}{x^{2}-1}$ continuous?

Question

$$\frac{x}{x^{2}-1}$$

a) $(-1,\infty )$

b) $(-\infty ,1)$

c)$(-1,1)$

d) none of the above

How can i solve this without using the graph?

Answer 1

For continuos functions $f(x)$ and $g(x)$ ; the function :

$$h(x)= \frac{f(x)}{g(x)}$$ is continuous everywhere except for the points where $g(x)=0$

The denominator is vanishing at at $x = \pm1 $. Therefore, the given function is discontinuous a t $x = \pm 1$.Hence, continuous in the interval : $(-\infty,-1) \cup (-1,1) \cup (1,\infty)$

Therefore the most correct answer is $\color{blue}{c. (-1,1)}$

Answer 2

Since $x^2-1=(x-1)(x+1),$ then the function is defined everywhere except $1$ and $-1.$

Answer a) includes $1$, so that cannot be correct. Answer b) includes $-1$ so that cannot be correct. Answer c) is suitable since it does not include $-1,$ or $1.$ However, the function is defined on $$\text{Domain}~=~(-\infty,-1)\cup (-1,1)\cup (1,\infty),$$ in total.

Answer c) is probably acceptable, although not the entire domain of this function.

Answer 3

Although the entire domain is not given in any option we know that, the value of x is not possible at 1 and -1 and the function becomes $\infty$ therefore the domain possible is $R$-{-1,1}