Question

lim x->0 ( 2xcos(1/x) + sin(1/x) )

Answer 1

Do you know L'Hopital's theorem?

Answer 2

yeah, i know a little... diff numerator n denominator ?

Answer 3

yes

Answer 4

how we use it here.. can we ?

Answer 5

Nope because when you take the limit and put 0 instead of x, you get 0 + 0, so the limit is 0

Answer 6

we apply L'Hopital when we get an indetermination like 0/0 or infinite / infinite

Answer 7

wait wait im wrong

Answer 8

you get 0·infinite

Answer 9

so yeah you have to apply L'Hopital

Answer 10

and transform it first to get 0/0 or infinite / infinite

Answer 11

To apply L'Hopital's rule, you must have it in a fraction form.

Answer 12

There is no indeterminate form. Why do you think there is? Both sine and cosine oscillate. They are not unbounded. The limit is not zero.

Answer 13

from that Q, how do u get lim x->0 ( 2xcos(1/x) + sin(1/x) ) ?

Answer 14

I'm solving it.

Answer 15

lim x->0 ( 2xcos(1/x) + sin(1/x) ) does not have a particular value.....

Answer 16

@hartnn
i diff (x^2)cos(1/x) ... because the Q ask.. is f'(x) cont at x=0 ...
am i wrong.... hm ?

Answer 17

x^2cos1/x or 2xcos 1/x?

Answer 18

all you can say is lim x->0 ( 2xcos(1/x) + sin(1/x) ) lies between +1 and -1

Answer 19

@GCR92
originally .. (x^2)cos(1/x)

Answer 20

in your pic, your just asked to take the derivative and check x=0 for continuity, not sure why your taking the limit

Answer 21

I don't understand the question now.

Answer 22

@hartnn wolf is helping :P

Answer 23

Use squeeze theorem for x^2cos(1/x)

Answer 24

@abb0t the limit does not exist :P

Answer 25

hba, what ?

Answer 26

@hartnn Nothing.

Answer 27

@MrDoe ... so dont have to use the limit ... ?
how to answer the question
' is f'(x) continous at x=0 ' ?

Answer 28

i think thats the confusion, the question says nothing about taking any limits

Answer 29

>.< im sorry

Answer 30

lim x->0 ( 2xcos(1/x) + sin(1/x) ) does not exist, so i'll say not continuous...

Answer 31

I'm trying to do the limit you said and it's a son of a...so I think MrDoe is right.

Answer 32

im sorry @GCR92 ..... thank you for your effort................

Answer 33

Np man

Answer 34

First, note:
-1 ≤ cos(x) ≤ 1
but you can ignore x = 0 because we are taking a limit and we know that limits don’t care about what’s actually going on at the point in question, x = 0 in the case.
Now if we have the above inequality for our cosine we can just multiply everything x squared. In other words we’ve managed to squeeze the function that we were interested in between two other functions that are very easy to deal with. So, the limits of the two outer functions are:
\[\lim_{x \rightarrow 0}x^2 = 0\]
\[\lim_{x \rightarrow 0}(-x^2) = 0\]
These are the same and so by Squeeze theorem we must also have
\[\lim_{x \rightarrow 0} x^2\cos(\frac{ 1 }{ x })=0\]

Answer 35

yeah, abbots on the right track just follow that

Answer 36

Graph: