The removable discontinuity of the function f(x) = (x+5)/(x^2 +3x-10) is at x = -5 (given by Option A).
Holes in the graph of function, where its undefined, or non-continuous, is called discontinuity.
The points in the domain of the function over which its not continuous, is called point of discontinuity for that function.
A discontinuity is removable if the limit of the function at the point of discontinuity exists but this limiting value is not the value of the function at that point.
We can remove that discontinuity by making the value of the function equate to the limiting value of the function at that point.
The given function is:
[tex]f(x) = \dfrac{(x+5)}{(x^2 +3x-10)}[/tex]
Factoring the denominator, we get:
[tex]x^2 + 3x -10 = x^2 + 5x - 2x - 10 =x(x+5) - 2(x+5) = (x+5)(x-2)[/tex]
Therefore, we get:
[tex]f(x) = \dfrac{(x+5)}{(x+5)(x-2)}[/tex]
The function is not defined if x = -5, or x = 2 since at those places, the denominator would become 0. (we cannot cancel out (x+5) from numerator and denominator for all x, as it isn't defined for x = -5
Also, we have:
[tex]\lim_{x\rightarrow 2}f(x)= \infty\\\lim_{x\rightarrow -5}f(x) = \lim_{x\rightarrow -5}\dfrac{(x+5)}{(x+5)(x-2)} = \lim_{x\rightarrow -5}\dfrac{1}{(x-2)} = -\dfrac{1}{7}[/tex]
We cancelled out (x+5) from numerator and denominator because x is limiting to -5 but isn't equal to -5.
For discontinuity of x = -5, the limit of f(x) exist (left and right limit both will come as 1/3). But f(-5) is not defined. So we can remove this discontinuity by defining f(-5) = 1/3 and for rest of the values of x, it is same as before.
Thus, the removable discontinuity of the function f(x) = (x+5)/(x^2 +3x-10) is at x = -5 (given by Option A).
Learn more about discontinuities here:
https://brainly.com/question/7327714
