Question

Evaluate \[\lim_{x \rightarrow 0^{+}} \frac{x^2sin\frac{1}{x}}{log(1+x)}\]

Answer 1

\[\lim_{x \rightarrow 0^{+}} \frac{x^2sin\frac{1}{x}}{log(1+x)}\]\[=ln10\lim_{x \rightarrow 0^{+}} \frac{2xsin\frac{1}{x}+x^2cos\frac{1}{x}(-\frac{1}{x^2})}{\frac{1}{1+x}}\]\[=ln10\lim_{x \rightarrow 0^{+}} (1+x)(2xsin\frac{1}{x}-cos\frac{1}{x})\]Hmmm...

Answer 2

lim x--->0 x sin(1/x)=0
and lim x--->0 ln(1+x)/x =1

Answer 3

so answer will be zero.

Answer 4

lim x--->0 x sin(1/x)=0 <- How to get this?

Answer 5

since -1 <= sin(1/x) < = 1

Answer 6

so lim x--->0 x sin(1/x)= 0

Answer 7

does it make a sense?

Answer 8

Not really.
-1 ≤ sinx ≤ 1
That's clear, but in this case, it's sin(1/x), where x can be 0.

Answer 9

x tends to 0+ so 1/x will tend to plus infinity.
so argument of sin will be infinity
but -1 <= sin(infinity)<=1

Answer 10

So my question is that is sin (infinity) valid?

Answer 11

yes, argument of sin can be infinity.
domain of sin is whole real no.

Answer 12

sin(infinity) will always have a value between -1 and +1.
So (sin1/x−cos1/x) will always have a value between -1 and +1.
Hence where is the problem you are facing

Answer 13

Then it would become \[\lim_{y \rightarrow \infty} siny\]Can we really evaluate this limit?

Answer 14

one minute...in ur solution ...u have applied hospitals rule ...right?

Answer 15

Applied once. Yes.

Answer 16

and this rule is applicable when the form is of 0/0 or infinity/infinity.

Answer 17

so ur denominator is 0 but how ur numerator is zero?

Answer 18

x^2... Hmm.. Perhaps that's where I got wrong..

Answer 19

just i mean that u r doing the same whatever i did if u r applying hospitals rule.

Answer 20

I say x^2, so directly put 0 into it to get 0/0.
Of course, I remember x sin (1/x) = 0, but I don't know/forget how to get that.

Answer 21

And what I've done doesn't mean that I'm correct..

Answer 22

x sin(1/x) is zeo if you expand sin(1/x).
http://upload.wikimedia.org/math/2/9/c/29cb648e96291f52707c7225630c1a17.png

Answer 23

I'm sorry but I haven't learnt expanding sinx.

Answer 24

can you graph xsin(1/x)

Answer 25

http://mathworld.wolfram.com/MaclaurinSeries.html
read this

Answer 26

see if the explanation here for lim x sin(1/x) = 0 helps
http://openstudy.com/users/waterineyes#/updates/5073f6e2e4b04aa3791e8596

Answer 27

sin(1/x) always lies between -1 and 1 and so as x approaches zero, 0<=|x sin(1/x)| <= |x| but since x->0, |x sin(1/x)| must approach zero also.

Answer 28

what kropot wrote is right

Answer 29

Sandwich Theorem! That's something I heard from my friends after the quiz!

Answer 30

Only problem is that I don't quite understand -1 </sin(1/x) </1 part..

Answer 31

magnitude of sine function cannot go beyond 1
sin 90 = 1 (max) sin 270 = -1 (min)
thats why those limits

Answer 32

*White flag* I don't understand what I understand.. :S
Thanks everyone!

Answer 33

Is infinity included the domain of sine?

Answer 34

*included in

Answer 35

yes. why not....but u cannot determine the value of sin(infinity)

Answer 36

hey, see the graph of sine function ..... its obvious

Answer 37

\[\lim_{x \rightarrow 0}=\frac{x ^{2}\sin \frac{1}{x}}{\log_{} (1+x)}=\frac{\frac{x ^{2}}{x}}{x}=1\]
the reasons being:
(1) as x ---> 0, sin 1/x = 1/x
(2) as x ---> 0, log (1 + x) = x

Answer 38

u cant say
lim x-->0 sin(1/x)= 1/x??

Answer 39

i think u r using for small x
sinx=x
but how u can approximate sin(1/x) with 1/x?

Answer 40

limit wont be 1, it'll be 0.

Answer 41

lim x--->0 x sin(1/x)=0
and
lim x--->0 ln(1+x)/x =1

so lim x-->0 x^2 sin(1/x)/ ln(1+x) =0

Answer 42

@akash123 Sorry. My bad :(

Answer 43

hey it's fine...:) plz no sorry.

Answer 44

best way to calculate simple limits is by using graphs....

Answer 45

Hmmm... Please ignore the following..
------------------------------------------------------------
sinx:
Domain: (-∞, ∞)
Range: [-1, 1]
So, -1 ≤ sinx ≤ 1 for (-∞, ∞)
-1 ≤ sin (1/x) ≤ 1
-x^2 ≤ x^2 sin (1/x) ≤ x^2

\[\lim_{x \rightarrow 0^+}-x^2 = 0\]\[\lim_{x \rightarrow 0^+}x^2 = 0\]By Sandwich Theorem,
\[\lim_{x \rightarrow 0^+}x^2 sin \frac{1}{x} = 0\]

For cosine
cosx:
Domain: (-∞, ∞)
Range: [-1, 1]
So, -1 ≤ cosx ≤ 1 for (-∞, ∞)
-1 ≤ cos (1/x) ≤ 1

How do you evaluate the limit of cos(1/x) when x-> 0+ ?

Answer 46

lim x-->0 cos(1/x)
indeterminate

Answer 47

take 1/x=y
then when y tends to infinity cos y tends to?
you yourself know this

Answer 48

yeah, thats inderterminate and u don't need it here

Answer 49

That's the problem!
\[=ln10\lim_{x \rightarrow 0^{+}} (1+x)(2xsin\frac{1}{x}-cos\frac{1}{x})\]
I can't evalute the cos(1/x)

Answer 50

\(\huge \lim_{x \rightarrow 0^{+}} \frac{xsin\frac{1}{x}}{log(1+x)/x}=\frac{0}{1}=0\)

Answer 51

distribute the limits in num, and denom

Answer 52

num limit through sandwich th.

Answer 53

How do you get that 1?

Answer 54

denominator lim through l'hopital
or u can use it as a standard formula

Answer 55

through L'hopital's its
1/(1+x) when x->0+ = 1

Answer 56

num derivative = 1/1+x denom derivative = 1, then at x=0+, the limit =1

Answer 57

.............. fraction in denominator...... can we still apply L'Hopital's rule for that? (Probably your answer is yes......)

Answer 58

aaaand it is yes

Answer 59

u just need to check for 0/0 in denominator

Answer 60

(Cry.....)
I am sooooooooo stupiddddd!!!
Thanks everyone!!!

Answer 61

\[\lim_{x \rightarrow 0^{+}} \frac{x^2sin\frac{1}{x}}{log(1+x)}\]
\[=\lim_{x \rightarrow 0^{+}} \frac{xsin\frac{1}{x}}{\frac{log(1+x)}{x}}\]
\[= \frac{\lim_{x \rightarrow 0^{+}}xsin\frac{1}{x}}{\lim_{x \rightarrow 0^{+}}\frac{log(1+x)}{x}}\]
Consider \(\lim_{x \rightarrow 0^{+}}xsin\frac{1}{x}\)
\[\lim_{x \rightarrow 0^{+}}xsin\frac{1}{x}\]
\[-1 \le sin\frac{1}{x} \le 1\]
\[-x \le xsin\frac{1}{x} \le x\]
\[lim_{x \rightarrow 0^+} –x = lim_{x \rightarrow 0^+} x = 0\]
By Sandwich Theorem,
\[lim_{x \rightarrow 0^+} xsin\frac{1}{x} = 0\]

Consider \(\lim_{x \rightarrow 0^{+}}\frac{log(1+x)}{x}\)
By L’Hopital’s Rule
\[\lim_{x \rightarrow 0^{+}}\frac{log(1+x)}{x}\]
\[=\lim_{x \rightarrow 0^{+}}\frac{\frac{1}{1+x}}{1}\]
\[=\lim_{x \rightarrow 0^{+}}\frac{1}{1+x} = 1\]

So,
\[\frac{\lim_{x \rightarrow 0^{+}}xsin\frac{1}{x}}{\lim_{x \rightarrow 0^{+}}\frac{log(1+x)}{x}}\]
\[=\frac{0}{1}=0\]

Answer 62

good work...:)

Answer 63

One minor question, when we do differentiation for common log, do we have to convert into natural log?

Answer 64

in questions relating limits, integration.... log is considered as natural log (ln)
unless its mentioned as log here is common log , in which case u need to convert it into natural log ln

Answer 65

so the answer to your question is YES

Answer 66

Awww... That means to be safe, I should convert it first, right?
(Since usually whenever I see log, it's common log)

Answer 67

you should convert only if it is mentioned that log is "common" log
else consider it as ln as you did it in this question
*only for problems on limits, diff., integration

Answer 68

Okay, thanks! So, for logx, there is no difference, but for \(\log_2x\), there is a problem, right?

Answer 69

yes, covert log2 x to ln x here

Answer 70

Nice! Thanks!!!~