Question

limits question

Answer 1

btw the 3 is not a part of the question. I accidently typed it in the box

Answer 2

Well, do you know about this identity:
\[\lim_{x \rightarrow 0}\frac{ sinx }{ x }=1\]?

Answer 3

yes

Answer 4

@Psymon

Answer 5

Alright, so as long as the angle of sin and the denominator match, the limit is 1. Even if its something crazy like sin(lnx)/lnx, since the angle and thedenominator match, the limit is 1.

Answer 6

i already tried that and it was wrong

Answer 7

So we want to manipulate this so we can get the form of sinx/x. So Ill rewrite what we have so we can visually see whats going on better:
\[\frac{ x^{2} }{ \sin(7x)\sin(7x) }\]

Answer 8

Well, hear me out then xD

Answer 9

okayy

Answer 10

So because we have sin(7x), the limit will be 1 if we can manage to get sin7x/7x. So Ill rewrite the problem we have once again and then show you what Ill do to accomplish that.

\[\frac{ \frac{ x^{2} }{ 1 } }{ \frac{ \sin(7x)\sin(7x) }{ 1 } }\]Again, this is just another change for visual sake.

Answer 11

right

Answer 12

So now what Ill do to get those sines to go away is ill multiply top and bottom by
\[\frac{ 1 }{ (7x)(7x) }\]
Doing this gives me:
\[\frac{ \frac{ x^{2} }{ (7x)(7x) } }{ \frac{ \sin(7x)\sin(7x) }{ (7x)(7x) } }\]So now that whole bottom is sin(7x)/7x, which is 1 times sin(7x)/7x which is also 1. So that whole sine portion I can just get rid of since when lim goes to 0, it all becomes 1. That leaves me with:

\[\frac{ x^{2} }{ (7x)(7x) }= \frac{ x^{2} }{ 49x^{2} }\implies \frac{ 1 }{ 49 }\]

Answer 13

ohhh okay..im just starting to learn this because i missed the lesson..thank you!

Answer 14

Yeah, np ^_^