Evaluate the limit of sqrt(x^3 + 4x^2)/x as x approaches 0, if it exists.

Step-by-step solution

1. 1

   Check the form. Both the radical and $x$ tend to $0$, so the quotient is $\frac{0}{0}$ and needs simplification.

2. 2

   Factor inside the radical. $x^3+4x^2=x^2(x+4)$, so $\sqrt{x^3+4x^2}=\sqrt{x^2}\sqrt{x+4}$ whenever $x+4\ge 0$.

3. 3

   Handle the square root correctly. The crucial point is $\sqrt{x^2}=|x|$, not $x$. Writing $\sqrt{x^2}=x$ is the mistake that makes this problem look as if the answer were $2$.

4. 4

   Reduce the quotient. $\frac{|x|\sqrt{x+4}}{x}=\operatorname{sgn}(x)\sqrt{x+4}$, since $\frac{|x|}{x}$ is $+1$ for $x>0$ and $-1$ for $x<0$.

5. 5

   Take the two one-sided limits. From the right, $\lim_{x\to 0^+}=+\sqrt{4}=2$. From the left, $\lim_{x\to 0^-}=-\sqrt{4}=-2$.

6. 6

   Conclude. The one-sided limits are finite but different, so the two-sided limit does not exist; the function has a jump of size $4$ at the origin.