Question

Calculating limits. (one small problem).

Answer 1

\[\lim_{t \rightarrow 0}\frac{ \sin3t }{ t }\]

Answer 2

\[=\lim_{t\to 0}3\frac{\sin(3t)}{3t}\]

Answer 3

Why do we put a 3 in front and how did the 3 appeared at the denominator?

Answer 4

\[=\lim_{t\to 0}\frac{\sin(3t)}{t}=\lim_{t\to 0}1\times\frac{\sin(3t)}{t}=\lim_{t\to 0}\frac{3}{3}\times\frac{\sin(3t)}{t}=\lim_{t\to 0}3\frac{\sin(3t)}{3t}\]

Answer 5

so whenever we have a limit that is undefined, such as this, we multiply it by 1?

Answer 6

the limit is defined

Answer 7

it can be computed

Answer 8

How is the limit defined?

Answer 9

Because, I'm used to seeing fractions that have a 0 at the denominator being undefined

Answer 10

the limit is of the form \(\dfrac{0}{0}\)

Answer 11

ah, I see.

Answer 12

making it an indeterminant form...which in this case the limit exists

Answer 13

Okay, got it now. Thank you :0

Answer 14

so what is the answer?

Answer 15

3

Answer 16

good

Answer 17

Just needed clarification on the concepts.