Question

limx->0 sin(abs(x))/x

If I am correct, than it is undefined? I mean... It isn't differentiable at all values because abs(x) isn't differentiable? So I can't do L'Hospitals??



@Mathematics

Answer 1

\[\lim_{x\rightarrow 0}\frac{\sin(|x|)}{x}\]?

Answer 2

think of it like this:

Case 1: x > 0 (so x approaches 0 from above):

= limx->0 sin(x)/x = 1

Case 2: x < 0 (so x approaches 0 from below):

= limx->0 (sin(-x))/x = limx->0 -sin(x)/x = -1

therefore the limit does not exist

Answer 3

yeah satellite

Answer 4

are you sure that limx->0+ of sin(x)/x =1??

Answer 5

jamesm nailed it. on one direction you get 1, in the other -1, no limit

Answer 6

it is always the case that
\[\lim_{x\rightarrow 0}\frac{\sin(ax)}{bx}=\frac{a}{b}\]

Answer 7

Does that work with all trig functions?

Answer 8

so
\[\lim_{x\rightarrow 0}\frac{\sin(x)}{x}=1\] and
\[\lim_{x\rightarrow 0}\frac{\sin(-x)}{x}=-1\]

Answer 9

And I can't apply L'hospital's to that sucker either right..?

Answer 10

yes, it's true that limx->0+ of sin(x)/x = 1

more generally sin(x)/x goes to 1 as x approaches 0 from both sides

Answer 11

no because it is not differentiable at 0

Answer 12

Okay, well what about this?

\[\lim_{x \rightarrow 0} (x-2)^{2}\sin(\pi/x-5) \]

Answer 13

\[l \lim_{x \rightarrow 5} (x-2)^{2} =9 \]

and

\[\lim_{x \rightarrow 5} \sin(\pi/x-5) = \sin(0) = 1\]

Which makes 1/0... So what do I do now?

Answer 14

limit as 0 or at 5?

Answer 15

this line
\[\lim_{x \rightarrow 5} \sin(\pi/x-5) = \sin(0) = 1\] is not correct.

Answer 16

pi/(x-5) or pi/x-5?

Answer 17

5

Answer 18

Oh crap. Okay so sin(pi/0) =???

No idea what to do!

Answer 19

division by zero is undefined

Answer 20

\[\lim_{x\rightarrow 5}\sin(\frac{\pi}{x-5})\] does not exist

Answer 21

\[\lim_{x \rightarrow 0} (x-2)^{2}\sin(\frac{\pi}{x-5}) \text{ is this the problem } \]

Answer 22

So the answer to this is undefined?

WolframAlpha is saying "Undefined on the interval (-9,9)". Is this what is expected of me on a quiz? where did they get the -9,9.

Answer 23

No myininaya, the limit of x is approaching 5

Answer 24

as x goes to 5, x - 5 goes to zero and
\[\frac{\pi}{x-5}\] goes to infinity. the sine therefore takes on all values between -1 and 1 infinitely often. there is no limit here

Answer 25

But wait! we have to multiply everything by 9. AHA, i get it!

Answer 26

Sweet, okay that makes sense. That's how WolframA got the interval (-9,9).

Thanks so much! that was a tricky one for me..

Answer 27

to be more precise
\[\frac{\pi}{x-5}\] goes to infinity if x > 5 and minus infinity if x < 5
in any case there is no limit.
the 9 is a red herring and has nothing to do with the problem

Answer 28

Okay, I'm glad you mentioned that last part. So I would not mention "on the interval (-9,9)" on a quiz.

Thanks!

Answer 29

So I would say on a quiz or something:

lim x->5 of sin(pi/x-5) is undefined, therefore
\[\lim_{x \rightarrow 5}(5-2)^{2}\sin(\pi/(x-5)\] is undefined as x is approaching 5?