Question

5.

Answer 1

Finding limits by direct substitution means simply means to evaluate the function at the desired value: in the first case, we have to evaluate $f(x)=x^2-3$ at $x=0$: we have

$f(0)=0^2-3 = 0-3=-3$

Similarly, in the second example, we have

$f(x)=x^2+3x-1 \implies f(3) = 3^2+3\cdot 3-1 = 9+9-1 = 17$

Going on, we have

$f(x) = \dfrac{x^2-16}{x-4} = \dfrac{(x+4)(x-4)}{x-4} = x+4$

And thus we have

$f(4) = 4+4=8$

Finally, we have

$f(x) = \dfrac{x^2-2x}{x^4} = \dfrac{x(x-2)}{x^4} = \dfrac{x-2}{x^3}$

So, we can't evaluate this function at 0.