Respuesta :
Finding limits by direct substitution means simply means to evaluate the function at the desired value: in the first case, we have to evaluate [tex]f(x)=x^2-3[/tex] at [tex]x=0[/tex]: we have
[tex]f(0)=0^2-3 = 0-3=-3[/tex]
Similarly, in the second example, we have
[tex]f(x)=x^2+3x-1 \implies f(3) = 3^2+3\cdot 3-1 = 9+9-1 = 17[/tex]
Going on, we have
[tex]f(x) = \dfrac{x^2-16}{x-4} = \dfrac{(x+4)(x-4)}{x-4} = x+4[/tex]
And thus we have
[tex]f(4) = 4+4=8[/tex]
Finally, we have
[tex]f(x) = \dfrac{x^2-2x}{x^4} = \dfrac{x(x-2)}{x^4} = \dfrac{x-2}{x^3}[/tex]
So, we can't evaluate this function at 0.
