# Limit questions

1. Apr 18, 2007

### transgalactic

i added a file in wich i tried to solved them

in every limit i solved plz correct me where i get it wrong.

maybe it some basic knowledge that i lack

plz help

#### Attached Files:

File size:
34.7 KB
Views:
66
• ###### p3.GIF
File size:
85.1 KB
Views:
47
2. Apr 18, 2007

### lalbatros

What is the meaning of lin(x)/x ???

3. Apr 18, 2007

### transgalactic

its S
sinx/x

lim (sinx)/x =1
x>0

4. Apr 18, 2007

### lalbatros

Then the first is easy.
It is of the form lim (f(x))^(sinx/x),
and lim f(x) is easyli obtained.

5. Apr 18, 2007

### transgalactic

so did i get it right??

6. Apr 19, 2007

### VietDao29

Nope, nope, it's incorrect.
You should note that the limit for e is:
$$\lim_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right) ^ x$$

Or

$$\lim_{x \rightarrow 0} \left( 1 + x\right) ^ \frac{1}{x}$$

All the two are of the Indeterminate Form $$1 ^ \infty$$, whereas in your problem, it's not any Indeterminate Forms.

--------------------

$$\lim_{x \rightarrow 0} \left( \frac{x ^ 2 - 2x + 3}{x ^ 2 - 3x + 2} \right) ^ \frac{\sin x}{x} = \left( \frac{0 ^ 2 - 2 \times 0 + 3}{0 ^ 2 - 3 \times 0 + 2} \right) ^ 1 = \frac{3}{2}$$

--------------------

The seconds problem you did it all correct, except for the last part, which should read:
$$= \fbox{\ln} e = 1$$

--------------------

The third problem, you've differentiated it incorrectly.

$$(\arctan (2x))' = \frac{2}{1 + 4x ^ 2}$$
$$(\sin (3x))' = 3 \cos (3x)$$

Ok, now, can you complete the three problems on your own? :)