Question

Anyone good with finding limits? Lim(x->0) x^2cos(1/x). Help is appreciated

Answer 1

L'hospital?

Answer 2

\[\lim_{x \rightarrow 0}\frac{\cos(\frac{1}{x})}{\frac{1}{x^2}}\]

Answer 3

Find derivative of the top
Find derivative of the bottom

Answer 4

doesnt that result in 1 regardless?

Answer 5

can you walk me through the derivative formula?

Answer 6

\[(\cos(\frac{1}{x}))'\]
Here you need to know the chain rule.

Answer 7

\[(\frac{1}{x})'=(x^{-1})' \]
Here you to know the power rule.

Answer 8

\[(\frac{1}{x^2})'=(x^{-2})'\]
Here you need to also know the power rule.

Answer 9

Chain rule goes like this:
\[(f(g(x))'=g'(x) \cdot f'(g(x))\]
Power rule goes like this:
\[(x^n)'=nx^{n-1} \]

Answer 10

so -2x^-2-1 so -2x^(-3) for the bottom?

Answer 11

\[\lim_{x \rightarrow 0}\frac{(\cos(\frac{1}{x}))'}{\frac{-2}{x^3}}\]
now we still need to take care of the top

Answer 12

\[(f(g(x))'=g'(x) \cdot f'(g(x))\]
\[(\cos(\frac{1}{x}))'=(\frac{1}{x})' \cdot (\cos)'(\frac{1}{x})\]

Answer 13

So what is the derivative of 1/x?

Answer 14

x^-1

Answer 15

yep what is the derivative of that

Answer 16

-1/x^2 dx = y' ?

Answer 17

right so we have
\[(f(g(x))'=g'(x) \cdot f'(g(x)) \]
\[(\cos(\frac{1}{x}))'=\frac{-1}{x^2} \cdot (\cos)'(\frac{1}{x}) \]

Answer 18

so we still need the derivative of cos

Answer 19

sin?

Answer 20

-sin

Answer 21

so we have
\[\lim_{x \rightarrow 0}\frac{\frac{1}{x^2}\sin(\frac{1}{x})}{\frac{-2}{x^3}} \]

Answer 22

is it -2/x^3 or -2/x^-3?

Answer 23

i guess i made a mistake with the -3, so based off that, is that the final solution then?

Answer 24

\[\frac{-1}{2} \lim_{x \rightarrow 0}\frac{ \frac{1}{x^2} \sin(\frac{1}{x})}{\frac{1}{x^2} \cdot \frac{1}{x}}\]

Answer 25

I'm totally lost on how that is simplified, arent you left with sin(1/x)/1/x, so -1/2* sin(0)?

Answer 26

1/x^2 cancels out with the 1/x^2 on bottom is how to simplify

Answer 27

but that leaves you with division by 0?

Answer 28

sin(1/x)/1/x= sin(1/0)/1/0

Answer 29

but \[\lim_{u \rightarrow \infty }\frac{\sin(u)}{u}=0\]

Answer 30

By squeeze thm.

Answer 31

so my initial answer of 0 was right even though i did it wrong lol?

Answer 32

yes -1/2*zero will be zero.

Answer 33

thank you i will print this out and review all the work, appreciate it

Answer 34

http://www.math.washington.edu/~conroy/general/sin1overx/
Here is a cute little visual showing you that x*sin(1/x) is between -x and x.
since -x and x both approach 0 as x approaches 0 then x*sin(1/x) approaches 0 as x approaches 0.

Answer 35

Someone may started out differently than I did here, but I just remember the rule that as u approaches infinity then sin(u)/u approaches 0.