# Finding Limit using L'Hopital's Rule

• Mar 29th 2009, 11:50 AM
suboba
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.
• Mar 29th 2009, 12:07 PM
Math Major
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.
• Mar 29th 2009, 12:24 PM
Chris L T521
Quote:

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