Question

Can someone please help me evaluate the limit below. I am very confused on what to do.

Answer 1

\[\lim_{x \rightarrow 0} \frac{ 1-2\cos x }{ 2x }\]

Answer 2

@uri could you possibly help?

Answer 3

Well what happens if you plug in 0?

Answer 4

it becomes 1/0

Answer 5

therefore we need to go furter to evaluate but im not very good with trig identities so i dont know the next step aka which trig identity i can use

Answer 6

hello?

Answer 7

nope you don't have to anything else
if you had 0/0 we would need go further

Answer 8

and actually you get -1/0
but still this means we are done and the limit is ?????

Answer 9

undefined?

Answer 10

Yeah
Or does not exist or (dne people say for short)

Answer 11

but that isn't how we learned in class. my teacher told us that we needed to use trig identities when we worked with practice problems.

Answer 12

\[\lim_{x \rightarrow 0}\frac{1-\cos(x)}{x}\]
if we had this one we have to go further because we do have 0/0

Answer 13

\[\lim_{x \rightarrow 0}\frac{1-\cos(x)}{x} \cdot \frac{1+\cos(x)}{1+\cos(x)} \\ \lim_{x \rightarrow 0}\frac{1-\cos^2(x)}{x(1+\cos(x)} \\ \lim_{x \rightarrow 0}\frac{\sin^2(x)}{x(1+\cos(x)} \\ \lim_{x \rightarrow 0}\frac{\sin(x)}{x} \frac{\sin(x)}{1} \frac{1}{1+\cos(x)} \\ \lim_{x \rightarrow 0} \frac{\sin(x)}{x} \cdot \lim_{x \rightarrow 0}\sin(x) \cdot \lim_{x \rightarrow 0} \frac{1}{1+\cos(x)} \\ (1)(0)(\frac{1}{1+1}) \\ (1)(0)(\frac{1}{2}) \\ 0 \]
But yeah we didn't have that function

Answer 14

ohhhh, okay. Thank you very much. I'm going to have to keep practicing these to get better. Thank you

Answer 15

if you had 0/0 you would definitely need those trig identities and recall some trig limits

Answer 16

Like the one I mentioned above...\[\lim_{x \rightarrow 0}\frac{\sin(x)}{x}=1 \text{ and } \lim_{x \rightarrow 0}\frac{1-\cos(x)}{x}\]

Answer 17

So do you think you would have to work further on this one...
\[\lim_{x \rightarrow 4}\frac{x+2}{x^2-16}\]?

Answer 18

like if we plug in 4 we would get 6/0
so we know the limit doesn't exist
we wouldn't have to bring out more algebra...

Answer 19

okay thank you so much!