Question

How do I find the limit of sin(2x) as x approaches pi?

Answer 1

@Melchizebek You could look at the graph of y = sin(2x) and determine the y-value when x = π.

In fact, you can do that without a graph. What is the value of sin(2π)?

Answer 2

0, right? Because Sin(π) is 0.

Answer 3

Yes. That is what I got.

Note: It is sin(2π) with a limit of 0 as x -->π
although sin(π) = 0 at that point also.

Answer 4

I see. I am having trouble with limits at the moment. With the one you just helped me with I was able to plug π in and get the answer. What about with

Answer 5

lim as x approaches 4 of cube root x+4?

Answer 6

\[\lim_{x \rightarrow 4}\sqrt[3]{x+4}\]

Answer 7

If you plug in 4 for this function, what do you get?

Answer 8

Oh! I tried that before, and was completely oblivious until just now that it equals two because the cube root of 8 is 2. So the limit as x approaches 4 of that function is 2.

Answer 9

I should have seen that myself. My bad.

Answer 10

These limits can be tricky.

Answer 11

So the first two works with plugging it in, but...
\[\lim_{x \rightarrow -3}x ^{2}+x-6/x+3\]

Answer 12

doesn't work with plugging it in, so how would I solve this one?

Answer 13

Would you put grouping symbols in the expression so that I'll know what is in a numerator and what is in a denominator? Thanks.

Answer 14

\[\lim_{x \rightarrow -3}(x ^{2}+x-6)\div (x+3)\]

Answer 15

Like that?

Answer 16

\[x ^{2}+x-6\]

Answer 17

is the numerator

Answer 18

and x+3 is the denominator

Answer 19

Here's how what you originally posted is correctly read without the grouping symbols.