Question

Calculating limits question

Answer 1

For \(\large{\epsilon > 0}\) evaluate the limit:

\(\large{\lim_{x \rightarrow \infty} x^{1-\epsilon} \int\limits_{x}^{x+1}\sin(t^2) dt}\)

Answer 2

I gave it some try :

First I am going to evaluate
\[\large{\int \sin(t^2) dt}\]

Answer 3

\[\large{\int \sin (t^2)dt }\]

\[\large{=\cfrac{-\cos(t^2)}{2t} - \cfrac{1}{2}\int\cfrac{\cos(t^2)}{t^2}dt}\]

Answer 4

L'hospital can be applied

Answer 5

\[\large \lim_{x \rightarrow \infty}\frac{ \int\limits_{x}^{x+1}\sin(t^2)dt}{ x^{\epsilon-1 }}\]

Answer 6

Now, for x > 0, I get,

\[\large{\int\limits_{x}^{x+1}\sin(t^2)dt = \cfrac{-\cos(x+1)^2}{2(x+1)} + \cfrac{\cos(x^2)}{2x} - \cfrac{1}{2}\int\limits_{x}^{x+1}\cfrac{\cos(t^2)}{t^2}dt}\]

Thus, I get:

\[\large{|\int\limits_{x}^{x+1}\sin(t^2)dt| \le \cfrac{1}{2(x+1)} + \cfrac{1}{2x} + \cfrac{1}{2}\int\limits_{x}^{x+1}\cfrac{dt}{t^2} = \cfrac{1}{x}}\]

Answer 7

Yeah I think I can use that too @sidsiddhartha

Answer 8

now we can differentiate by leibnitz rule

Answer 9

Yeah

Answer 10

yeah ur process is more mathematical :)

Answer 11

But coming back to my method:

Thus, I can say:

\[\large{|x^{1-\epsilon}\int\limits_{x}^{x+1}\sin(t^2)dt| \le \cfrac{1}{x^{\epsilon}}}\]

Now, since, as \(\large{x \rightarrow \infty}\), \(\large{\cfrac{1}{x^{\epsilon}} \rightarrow 0}\), thus, we can deduce that:

\(\large{\lim_{x \rightarrow \infty} x^{1-\epsilon} \int\limits_{x}^{x+1}\sin(t^2) dt = 0}\)

Answer 12

I always used LH Rule for limits problems, so this time I thought to follow some different method.

Answer 13

well done :)

Answer 14

thanks @sidsiddhartha :)

Answer 15

This user rocks ^ ^ ^ ... He learnt this all from me.. (just to let u all know ) lol

Answer 16

Is there some third method also friends ?

Answer 17

Learning alternative methods always help :)

Answer 18

bhaiya se aise baat karte hai O.o @mathslover

Answer 19

Okay lets give it a try by LH Rule:

\(\color{blue}{\text{Originally Posted by}}\) @sidsiddhartha
\[\large \lim_{x \rightarrow \infty}\frac{ \int\limits_{x}^{x+1}\sin(t^2)dt}{ x^{\epsilon-1 }}\]
\(\color{blue}{\text{End of Quote}}\)

[@sidsiddhartha , OS pe bhaiya se aise hi baat karte hain ;) ]

Answer 20

Haha.. yeah, reality mein dar lagta hai bhaiya se "aise" baat karne mein.. but OS par chalta hai..

Answer 21

Okay , I am going to apply LH Rule:

So, I have,

\[\large{\cfrac{ \int\limits_{x}^{x+1}\sin(t^2)dt}{ x^{\epsilon-1 }}}\]

Applying LH Rule,

\[\large{\cfrac{\sin(x+1)^2 - \sin(x^2)}{(\epsilon -1)x^{\epsilon-2}}}\]

Now, what to do next @sidsiddhartha ?

Any ideas ? We are not given whether \(\epsilon\) is more than 1. So, we can't say that \(\large{x^{\epsilon - 2}}\) will be positive or negative.