# Thread: L'Hopital's Rule - finding limit problem

1. ## L'Hopital's Rule - finding limit problem

I wanted to make sure I'm doing this right.
1. lim x--> 0. $\frac{x^3}{sinx-x}$
Get 0/0, so can use L'hopital's rule.
$\frac{3x^2}{cosx-1}$= $\frac{-6x}{sinx}$= $\frac{-6}{cosx}$ Plugging a 0 into the last one gives -6. Is this technique right?

2. lim x--> infinity. $\frac{x^2+4x}{9x^3+4}$.

I'm confused on how to approach an infinity one. How do I see both the numerator and denominator go to infinity(well, I guess I can tell if I graph). Can someone walk me through the steps? I would really appreciate it.

2. 1. Remember that the derivative of cosine is negative sine, so your result should be positive.

2. If x goes to infintiy then both the numerator and the denominator go to infinty because they both contain x. It has nothing to do with the graph of the function. If you substitute the infinity for x, they both go to infinity. So just differentiate using L'hopitals rule. It should converge to zero.

3. Originally Posted by bcahmel
I wanted to make sure I'm doing this right.
1. lim x--> 0. $\frac{x^3}{sinx-x}$
Get 0/0, so can use L'hopital's rule.
$\frac{3x^2}{cosx-1}$= $\frac{-6x}{sinx}$= $\frac{-6}{cosx}$ Plugging a 0 into the last one gives -6. Is this technique right?
Find $\displaystyle \lim_{x\to 0}\frac{3x^2}{\cos x-1}$

Originally Posted by bcahmel
2. lim x--> infinity. $\frac{x^2+4x}{9x^3+4}$.
Try

$\displaystyle\frac{x^2+4x}{9x^3+4}\times \frac{\frac{1}{x^3}}{\frac{1}{x^3}}$

4. Originally Posted by bcahmel
I wanted to make sure I'm doing this right.

2. lim x--> infinity. $\frac{x^2+4x}{9x^3+4}$.

I'm confused on how to approach an infinity one. How do I see both the numerator and denominator go to infinity(well, I guess I can tell if I graph). Can someone walk me through the steps? I would really appreciate it.
This can be done without Lhopital's. Divide the numerator and denominator by $x^3$ and proceed.

5. Originally Posted by bcahmel
I wanted to make sure I'm doing this right.

1. $\displaystyle\lim_{x \to\ 0} \frac{x^3}{sinx-x}$

Get 0/0, so can use L'hopital's rule.

$\displaystyle\frac{3x^2}{cosx-1}\rightarrow\frac{-6x}{sinx}\rightarrow\frac{-6}{cosx}$

Plugging a 0 into the last one gives -6. Is this technique right? correct

2. $\displaystyle\lim_{x \to \infty}\frac{x^2+4x}{9x^3+4}$.

I'm confused on how to approach an infinity one. How do I see both the numerator and denominator go to infinity(well, I guess I can tell if I graph). Can someone walk me through the steps? I would really appreciate it.
$\displaystyle\lim_{x \to \infty}\frac{x^2+4x}{9x^3+4}=\lim_{x \to \infty}\frac{x^3\left(\frac{1}{x}+\frac{4}{x^2}\ri ght)}{x^3\left(9+\frac{4}{x^3}\right)}$

2. So if I divide by x^3, I get: $\frac{x^-1+4x^-2}{9+4x^-3}$, is that right? This way is easier than L'Hopital's Rule? because if I did it that way I would get $\frac{2x+4}{27x^2}$ but I guess I would have to take more derivatives of this... This is confusing!
2. So if I divide by x^3, I get: $\frac{x^-1+4x^-2}{9+4x^-3}$, is that right? This way is easier than L'Hopital's Rule? because if I did it that way I would get $\frac{2x+4}{27x^2}$ but I guess I would have to take more derivatives of this... This is confusing!
$\displaystyle\frac{2}{54x}$