Question

Lim (xlnx +2xln(sin1/√x) )
where x---->infinity

Answer 1

Is this the limit?
\[
\lim_{x\to\infty}\left[x\ln(x)+2x\ln\left(\sin\left(\frac{1}{\sqrt{x}}\right)\right)\right]
\]

Answer 2

the limit does not exist :))))

Answer 3

The limit might exist and is infinity...

Answer 4

Mathematica returned a surprising answer.... The limit exists and is finite?!

Answer 5

@ganeshie8 Help!
\[
\begin{align*}
&\phantom{{}={}}\lim_{x\to\infty}\left[x\ln(x)+2x\ln\left(\sin\left(\frac{1}{\sqrt{x}}\right)\right)\right]\\
&=\lim_{x\to\infty}\left[\ln(x^x)+\ln\left(\sin\left(\frac{1}{\sqrt{x}}\right)^{2x}\right)\right]\\
&=\lim_{x\to\infty}\ln\left(x^x\sin\left(\frac{1}{\sqrt{x}}\right)^{2x}\right)\\
&=\lim_{x\to\infty}\ln\left(\left(x\sin^2\left(\frac{1}{\sqrt{x}}\right)\right)^x\right)\\
\end{align*}
\]
\[
\begin{align*}
&\phantom{{}={}}\lim_{x\to\infty}\left(x\sin^2\left(\frac{1}{\sqrt{x}}\right)\right)^x\\
&=\lim_{x\to\infty}\left(\sqrt{x}\sin\left(\frac{1}{\sqrt{x}}\right)\right)^{2x}\\
\end{align*}
\]

Answer 6

On what basis did you remove the logarithmic sign in the second last step?

Answer 7

Since \(\ln\) is a continuous function, we could move the limit inside and then take the logarithm.

Answer 8

As in \[
\lim_{x\to a}\ln\left(f(x)\right)=\ln\left(\lim_{x\to a}f(x)\right)
\]

Answer 9

Ah! Ohk..i didn't see a logarithm sign outside the limit.That's why asked if it has been removed or not.

Answer 10

This is 1^infinity limit now.

Answer 11

My doubt resides in this, that one method gave me the and and other didn't. I want to know the fault in the other method.
@thomas5267 will you pls help?

Answer 12

I used expansion in first method and in the second I used l'hospital rule .It didn't gave me the and though.

Answer 13

One method gave you what?

Answer 14

-1/3

Answer 15

-1/3 is the correct answer but I don't know how to get it.

Answer 16

I type the original limit into Mathematica and -1/3 was the answer. How did you get that answer though?

Answer 17

We need to apply expansion for the sin function

Answer 18

Sine*

Answer 19

We need to apply this.
e^((√xsin(1/√x)-1)*2x)
Apply expansion for sine function and when we solve the limit we will get -1/3.

Answer 20

Can't we apply L'hospital rule here? I got limit to be infinity then .

Answer 21

How to L'hospital here? I don't see any straight forward way.

Answer 22

We can apply it bcoz we can move that 2x term in the denominator and then we would be have 0/0 form.

Answer 23

Don't really understand how do you move it to denominator.

Answer 24

We can write like this.
(√xsin(1/√x)-1)/(1/2x)
Now it's a 0/0 form.

Answer 25

(√xsin(1/√x)-1)/(1/2x)

Where did you get this from? I don't see how this corresponds to anything in the calculation.

Answer 26

Why not? Earlier also we had the same same expression with 2x in numerator.
Now i just shifted that in denominator as 1/2x.

Answer 27

I am pretty sure the two expressions are not equivalent. You cannot combine logarithms with coefficients in front of them. You have to convert the coefficient to power inside the logarithm before combining them.

Answer 28

Could you show your steps? I can't figure out how you get that expression.

Answer 29

Sure...
Did you get that lim that we need to evaluate is (√xsin(1/√x)-1)2x

Answer 30

No...

Answer 31

Where is the difficulty in getting it?

Answer 32

The limit we need to evaluate is \[
\lim_{x\to\infty}\left(\sqrt{x}\sin\left(\frac{1}{\sqrt{x}}\right)\right)^{2x}
\]
Where did you get that -1 from or am I looking at the completely wrong thing?

Answer 33

This is 1^infinity limit.Right?

Answer 34

Yes.

Answer 35

How do we evaluate them?
Like say you have f(x)^g(x)
Where f(x)----->1 and g(x)---->infinity then how you solve them?

Answer 36

I don't think L'Hospital can be applied here.

Answer 37

It is not a 0/0 limit.

Answer 38

I don't know how to solve it but I think series expansion is the only method.

Answer 39

Why not 0/0 form?
Pls show that

Answer 40

\[
\lim_{x\to\infty}\left(\sqrt{x}\sin\left(\frac{1}{\sqrt{x}}\right)\right)^{2x}
\]How can this be converted to 0/0?

Answer 41

I told you already.Please suggest a way as to why is it not 0/0 form.
I gave you the evidence also that the limit is (√xsin(1/√x)/(1/2)x)

Answer 42

Forgot to put -1 in the numerator

Answer 43

I don't see how you could convert it to 0/0 form is the problem. Yes \(\left(\sqrt{x}\sin\left(\frac{1}{\sqrt{x}}\right)-1\right)/\dfrac{1}{2x}\)is a 0/0 form but I don't understand where did you get that from. You might have screwed up some step and hence the answer is incorrect.

Answer 44

I got this from here .
If we have lim (1+f(x))^g(x) with f(x) tends to 0 and g(x) tends to infinity then we always solve this using e^f(x)g(x) .
I applied the same here also and i got the result you wrote above

Answer 45

\[
\frac{\sqrt{x}\sin\left(\frac{1}{\sqrt{x}}\right)-1}{\frac{1}{2x}}
\]
I don't think L'Hospital will work here since the denominator is \(2x^{-1}\). The n-th derivative of \(2x^{-1}\) is \((-1)^n(2)(n!)x^{-n-1}\) and the limit to infinity of this is always 0.

Answer 46

Does L'hospital rule fail if we have limit to be 0?

Answer 47

The problem is that the limit of the derivatives of both denominator and numerator are 0. You can not find the limit easily.

Answer 48

No matter how many times you differentiate the denominator, the limit to infinity of the denominator is still 0. If the limit to infinity of derivatives of numerator is not 0 then you can rightfully conclude that the limit blows up to infinity. The problem is you apply L'Hospital once and you get 0/0 again. I applied L'Hospital 4 times on Mathematica and it is still 0/0.

Answer 49

Okay ..thanks and from now i will prefer to use the expansion bcoz i have learnt that expansion never give you dubious results.

Answer 50

It is not so much of a dubious result but since you apply L'Hospital to the limit you get 0/0 you have to apply it again.

Answer 51

And by the fourth iteration of L'Hospital you know that this is going nowhere lol.

Answer 52

Yes! The same for me.I find it literally sick if it exceeds second iteration of L'hospital for ugly expressions (particularly)

Answer 53

There is no way L'Hospital can work here since the limits to infinity of derivatives of denominator are always 0. There is no chance that you can get -1/3.

Answer 54

And I tried the limits to infinity of the derivatives of numerator using Mathematica. By the 30th iteration the derivative is still 0 lol.

Answer 55

That's sickk thing..
Lol.
Bye too late now.and thanks agian.