Question

FAN AND MEDAL

Answer 1

@geerky42

Answer 2

@hedyeh99

Answer 3

@waterineyes

Answer 4

Firstly:

\[\sin(-x) = -\sin(x)\]

Answer 5

http://www.wolframalpha.com/input/?i=lim+%28sin%28-4x%29%2Fsin%28-2x%29%29

Answer 6

\[\lim_{x \to 0}\frac{\sin(4x)}{\sin(2x)}\]

Answer 7

Secondly:

\[\sin(2A) = 2 \sin(A) \cos(A)\]

Answer 8

Use that formula to only numerator part.. :)

Answer 9

thank you!

Answer 10

can you help me with my last one?

Answer 11

\[\lim_{x \to 0}\frac{2\sin(2x) \cos(2x)}{\sin(2x)} \implies \lim_{x \to 0}2 \cos(2x)\]

Put x = 0, here and you get limit as 2..

Answer 12

There use:

\[a^2 - b^2 = (a+b)(a-b)\]

Answer 13

You will see \((x+4)\) getting cancelled there.. :)

Answer 14

\[f(x) = \frac{(x+4)(x-4)}{x+4} \implies f(x) =x-4 \; \quad Put \; x = -4 \; here\]

Answer 15

What did you get when you put x = -4 there?

Answer 16

To understand why, clearly \(\dfrac{x^2-16}{x+4}\) is discontinuous at \(x=-4\), but since it said it is continuous at \(x=-4\), then \(f(-4)\) must be equal to \(\displaystyle\lim_{x\rightarrow-4}\dfrac{x^2-16}{x+4}\), which can be solve by doing what @waterineyes said.

Does that make sense?

Answer 17

-8

Answer 18

Good..!!

Answer 19

yay(:

Answer 20

Nice @geerky42 ..!!