# Thread: Finding Limit using L'Hopital's Rule

1. ## Finding Limit using L'Hopital's Rule

How do you find these limits using L'Hopital's rule?

1. $\lim_{x \to 0}~\frac{sin x} {x^{1/3}}$

2. $\lim_{x \to 0}~\frac{x} {(sin x)^{1/3}}$

In both cases, when I apply L'Hopitals rule, the derivative of the denominator won't allow me to solve because there ends up being a zero in there. ie $1/3x^{-2/3}$

Thanks.

2. In both cases, you should have a negative power in the denominator which moves it up to the numerator. Be sure that you simplify the expression before evaluating.

3. Originally Posted by suboba
How do you find these limits using L'Hopital's rule?

1. $\lim_{x \to 0}~\frac{sin x} {x^{1/3}}$

2. $\lim_{x \to 0}~\frac{x} {(sin x)^{1/3}}$

In both cases, when I apply L'Hopitals rule, the derivative of the denominator won't allow me to solve because there ends up being a zero in there. ie $1/3x^{-2/3}$

Thanks.
You can actually do these without L'Hopitals Rule (but you need to know that $\lim_{x\to0}\frac{\sin x}{x}=1$ ):

1. $\lim_{x\to0}\frac{\sin x}{x^{\frac{1}{3}}}=\lim_{x\to0}x^{\frac{2}{3}}\cd ot\frac{\sin x}{x}=\lim_{x\to 0}x^{\frac23}\cdot\lim_{x\to0}\frac{\sin x}{x}=0\cdot1=\boxed{0}$

2. $\lim_{x\to0}\frac{x}{\left(\sin x\right)^{\frac{1}{3}}}=\lim_{x\to0}\left(\sin x\right)^{\frac23}\cdot\frac{x}{\sin x}=\lim_{x\to0}\left(\sin x\right)^{\frac{2}{3}}\cdot\lim_{x\to0}\frac{x}{\s in x}$ $=\lim_{x\to0}\left(\sin x\right)^{\frac{2}{3}}\cdot\frac{1}{\lim\limits_{x \to0}\displaystyle\frac{\sin x}{x}}=0\cdot\frac{1}{1}=\boxed{0}$