1. ## Limits.

"Find the limit or explain why it does not exist"

I'm having trouble with these problems, I can't factor/these into a different form.
Having trouble with the algebra.
I've just started calculus and we haven't gone very far so we've just covered limit laws.

I do understand for number 7 that "lim sin(x)/x = 1", but lim sin(2x)/(2x + 3x) doesn't seem right.

2. ## Re: Limits.

Originally Posted by Zirrick
"Find the limit or explain why it does not exist"

I'm having trouble with these problems, I can't factor/these into a different form.
Having trouble with the algebra.
I've just started calculus and we haven't gone very far so we've just covered limit laws.

I do understand for number 7 that "lim sin(x)/x = 1", but lim sin(2x)/(2x + 3x) doesn't seem right.
5. Get a common denominator so that you can perform the subtraction, then divide by x. You should be able to substitute x = 1 then.

7.
\displaystyle \begin{align*} \frac{\sin{2x}}{5x} &= \frac{1}{5}\left(\frac{\sin{2x}}{x}\right) \\ &= \frac{2}{5}\left(\frac{\sin{2x}}{2x}\right) \end{align*}

Now make use of the fact that \displaystyle \begin{align*} \lim_{X \to 0}\frac{\sin{X}}{X} = 1 \end{align*}.

8. This is a perfect candidate for L'Hospital's Rule.

9. Since logarithms are only defined for positive values of the independent variable, is it possible to approach 0 from the left? What does this tell you about this problem?

11. Divide top and bottom by \displaystyle \begin{align*} x^3 \end{align*}.

3. ## Re: Limits.

For #7:

limit sin(2x)/(5x) = limit sin(2x)/2x * limit 2x/5x = 1 * 2/5 = 2/5

Just picture it.. if the denominator were equal to (2x), the limit would be 1, but instead, it is larger than that by a scale of (5/2)... therefore you need to divide by that much extra, to get 2/5.

For #8, the same principle applies actually. As x -> 0 in this case, you can treat tan() as sin(), because you are dividing sin() by a quantity that approaches 1, cos(x). So the answer is 2/Pi.

For #5, multiply both the numerator and the denominator of the big fraction by 2(2+x). It should then simplify nicely.

4. ## Re: Limits.

Hello, Zirrick!

Here is #8 without using L'Hopital . . .

$8.\;\lim_{x\to0}\frac{\tan(2x)}{\tan(\pi x)}$

$\frac{\tan(2x)}{\tan(\pi x)} \;=\;\dfrac{\frac{\sin(2x)}{\cos(2x)}}{\frac{\sin( \pi x)}{\cos(\pi x)}} \;=\; \frac{\sin(2x)}{1}\cdot\frac{1}{\sin(\pi x)}\cdot\frac{\cos(\pi x)}{\cos(2x)}$

. . $=\;{\color{red}\frac{2x}{2x}}\cdot\frac{\sin(2x)}{ 1}\cdot {\color{red}\frac{\pi x}{\pi x}}\cdot\frac{1}{\sin(\pi x)} \cdot \frac{\cos(\pi x)}{\cos(2x)} \;=\;\frac{2x}{\pi x}\cdot\frac{\sin(2x)}{2x}\cdot\frac{\pi x}{\sin(\pi x)}\cdot\frac{\cos(\pi x)}{\cos(2x)}$

$\lim_{x\to0}\left[\frac{2}{\pi}\cdot\frac{\sin(2x)}{2x}\cdot\frac{ \pi x}{\sin(\pi x)}\cdot\frac{\cos(\pi x)}{\cos(2x)}\right] \;=\; \frac{2}{\pi}\cdot1\cdot1\cdot\frac{1}{1} \;=\;\frac{2}{\pi}$