Question

lim x approaches zeros from the right of (sinx/5*sqrt(x))

SInce zero makes the denominator zero I put zero in a sin chart. I chose 1 and plugged it in to get positive infinity. but my book says the answer is 0. can someone explain why

Answer 1

\[\lim_{x \rightarrow 0}\frac{\sin(x)}{5 \sqrt{x}}\]

Answer 2

"since 0 makes the denominator =0" but you CAN'T assign another number to calculate it. :)

Answer 3

yes from the right

Answer 4

use that one limit
x->0 of sin(x)/x=1
first multiply top and bottom by sqrt(x)

Answer 5

so is it to rationalize it? and then since its the graph of a square root?

Answer 6

I suppose you are asking why I'm multiplying top and bottom by sqrt(x) ? This is because I want to attempt using as x approaches 0 sin(x)/x approaches 1 .

Answer 7

\[\lim_{x \rightarrow 0} \frac{ \sqrt{x}}{5} \frac{\sin(x)}{x}\]

Answer 8

and I mean to put from the right

Answer 9

ok I got it but is it zero because its the graph of a square root or is it because

\[\frac{ \sqrt{x} }{ 5 }* \frac{ sinx}{ x }\] = 0 * 1

Answer 10

The function doesn't exist to the left of 0 because of the square root
But the function does exist to the right of 0 and as we approach 0 from the right we have 0/5 * 1 =0*1=0
yes

Answer 11

thank you freckles youre the best!

about that sinx/x =1 theorm.
would it also work for sinx/x^2?

Answer 12

\[\lim_{x \rightarrow 0}\frac{\sin(x^2)}{x^2}=1 \\ \lim_{x \rightarrow 1}\frac{\sin(x-1)}{x-1}=1 \\ \lim_{x \rightarrow 0}\frac{\sin(x^5)}{x^5}=1 \\ \text{ but } \\ \lim_{x \rightarrow 0}\frac{\sin(x)}{x^2}=\lim_{x \rightarrow 0} \frac{\cos(x)}{2x}=\frac{1}{0} \text{-> this limit does not exist }\]

Answer 13

nice it does not exist but goes to positive infinity right

Answer 14

I would say this but if u(x)=sqrt(x) then the limit will only exist to the right ...
\[\lim_{u(x) \rightarrow 0}\frac{\sin(u(x))}{u(x)}=1 \]
so this sorta is what i'm saying
with some restrictions on u(x)

Answer 15

actually it depends on what side you are talking about

Answer 16

sin(x)/x^2 is actually an odd function
since sin(-x)/(-x)^2=-sin(x)/x^2

Answer 17

so that means the left and right limit of 0 will be doing opposite things

Answer 18

one side will go to pos inf the other will go to neg inf

Answer 19

we know sin(x)/x^2 to the right of 0 will definitely go to pos inf because both sin and x^2 are pos for values close to 0 on right
however on the left of 0, sin(x)/x^2 will go to neg inf because sin will be neg for values close to 0 on the left

Answer 20

and x^2 is always positive for either direction
sin(x) is what determines the neg of pos values

Answer 21

yes I was just talking sbaout apr the right. but thank you for answering all my questions!

Answer 22

np