# Thread: How do I solve these?

1. ## How do I solve these?

I tried them several times and got them wrong. I don't understand the way my prof explains how to do them either.

If you could show step by step, as simple as possible, so that I can understand and be able to do them myself.

$\lim_{x \to 0}\frac{1-cosx}{x³}$ lim does not exist

and

$\lim_{x \to 0}\frac{tanx}{sin8x}$ lim is $\frac{1}{8}$

$\lim_{x \to 0} 7x²\cdot cotx\cdot cscx$ lim is $\frac{7}{11}$

I already have the correct answer - I just don't understand how to get there.

2. Originally Posted by ImaCowOK
I tried them several times and got them wrong. I don't understand the way my prof explains how to do them either.

If you could show step by step, as simple as possible, so that I can understand and be able to do them myself.

Limit of $\frac{1-cosx}{x³}$ as x -> 0 lim does not exist

and

Lim of $\frac{tanx}{sin8x}$ as x -> 0 lim is $\frac{1}{8}$

I already have the correct answer - I just don't understand how to get there.
$\displaystyle \frac{1-\cos{x}}{x^3} \cdot \frac{1+\cos{x}}{1+\cos{x}} = \frac{1-\cos^2{x}}{x^3(1+\cos{x})} = \frac{\sin^2{x}}{x^3(1+\cos{x})} = \frac{\sin^2{x}}{x^2} \cdot \frac{1}{1+\cos{x}} \cdot \frac{1}{x}
$

now evaluate ...

$\displaystyle \lim_{x \to 0} \frac{\sin^2{x}}{x^2} \cdot \frac{1}{1+\cos{x}} \cdot \frac{1}{x}$

for the second limit ...

$\displaystyle \frac{\tan{x}}{\sin(8x)} \cdot \frac{8x}{8x} = \frac{8x \tan{x}}{8x \sin(8x)} = \frac{1}{8} \cdot \frac{\tan{x}}{x} \cdot \frac{8x}{\sin(8x)}$

now evaluate ...

$\displaystyle \lim_{x \to 0} \frac{1}{8} \cdot \frac{\tan{x}}{x} \cdot \frac{8x}{\sin(8x)}$

3. Originally Posted by skeeter
$\frac{\tan{x}}{x}$
How did you get that? I got $\frac{\tan{x}}{8x}$ instead on that step after $= \frac{8x \tan{x}}{8x \sin(8x)}$

4. Originally Posted by ImaCowOK
How did you get that? I got $\frac{\tan{x}}{8x}$ instead on that step after $= \frac{8x \tan{x}}{8x \sin(8x)}$
$\displaystyle \frac{\tan{x}}{8x} = \frac{1}{8} \cdot \frac{tan{x}}{x}$

5. Okay, I get it.