Limit as $x$ approaches to $0$ is nonzero divided by $0$

Question

The equation I am dealing with is:

$$\lim_{x \to 0} \frac{x^2+3x}{3x^5}$$

First I tried to use L'Hopital's Rule but I got stuck with an equation $\frac{2x+3}{15x^4}$. Another approach to this equation was to cancel $x$: $\frac{x+3}{3x^4}$, but it also gives nothing.

The right answer is $0$.

Answer 1

Note that following your way

$$2x+3\to 3$$

$$15x^4\to 0^+$$

therefore

$$\lim_{x \to 0} \frac{x^2+3x}{3x^5}\stackrel{H.R.}=\lim_{x\to0}\frac{2x+3}{15x^4}=+\infty$$

Note also that we don't need l'Hopital indeed we can simply cancel out a $x$ factor from numerator and denominator to obtain

$$\lim_{x \to 0} \frac{x^2+3x}{3x^5}=\lim_{x \to 0} \frac{x+3}{3x^4}=+\infty$$

indeed $x+3\to 3$ and $3x^4\to 0^+$.

Answer 2

Why did you get stuck when you got $\displaystyle\lim_{x\to0}\frac{2x+3}{15x^4}$? This limit is equal to $+\infty$. Therefore, the limit that you're after is $+\infty$ too.