Question

am i solving this correctly? limx->0 sin 7x/4x

Answer 1

\[\lim_{x \rightarrow 0} \frac{ \sin7x }{ 4x }\]

Answer 2

\[\lim_{x \rightarrow 0} \frac{ 4 }{ 7 }( \frac{ \sin x }{ x })\]

Answer 3

\[\frac{ 4 }{ 7 }\lim_{x \rightarrow 0} \frac{ \sin x }{ x }\]

Answer 4

sin x/x is just 1 and 4/7 times 1 is just 4/7

Answer 5

done?

Answer 6

\[\lim_{x\to0}\frac{\sin ax}{ax}=1\]
So,
\[\lim_{x\to0}\frac{\sin7x}{4x}=\frac{7}{7}\lim_{x\to0}\frac{\sin7x}{4x}=\frac{7}{4}\lim_{x\to0}\frac{\sin 7x}{7x}=\cdots \]

Answer 7

The answer should be the reciprocal of what you got: 7/4.

Answer 8

haha yes, ofcourse, typo there. i meants 7/4

Answer 9

meant*

Answer 10

so

Answer 11

\[\lim_{x \rightarrow 0} \frac{ 7 }{ 4 }( \frac{ \sin x }{ x })\]

Answer 12

\[\frac{ 7 }{ 4 }\lim_{x \rightarrow 0} \frac{ \sin x }{ x }\]

Answer 13

= 7/4

Answer 14

No, multiply by 7/7 (=1) and then switch out the 4x for 7x, then the sin7x or 7x turns into one. 1 times 7/4 will be 7/4. You got the right answer, but in the wrong way.

Answer 15

you just switch.. ok then