evaluate lim x→0 x sin 1/x
Answers
well this question might seem an easy one but it isn't actually easy,at least not at all easy until you have aced LCD and have the idea about the range of the various trigonometric functions. Actually the angle inside sin i.e (1/x) makes this question a bit tricky to tackle. Why so? Let's see.
If you substitute the limiting value in the expression,
lim x → 0 (0) sin (1/0)
lim x → 0 (0) sin (∞)
And we are aware that the range of sin is [-1,1] so sin (1/x) will actually oscillate between -1 to +1,this happens because in limits we are dealing with tending to values and not the exact values and infinity is not some number,so we have no clue about what angle are we talking about,so it opens the door to the entire range of sin. Like you put any number instead of x in sin (x) the value will surely lie somewhere from -1 to 1.
So,how do we proceed further? You can use either the intuitive way to proceed or the conceptual way. Let me do the both ways for you since the solution isn't much lengthy.
The intuitive way or you can using the basics of limit :
So, substitute the limiting value first of all,
lim x → 0 x sin (1/x)
(→ 0) × sin (1/→0)
(→0) × (this will oscillate from -1 to 1}
So,now here onwards you can think in two ways,the first way is that if the sin function will keep oscillating from -1 to 1,it will make the limit invalid,so there's no chance of limit existing, so asking this question is a very vague idea,since you have been asked to *evaluate*.
Hence to make the limit exist and finite,the only possibility is the value must become zero, that's the only way left to get rid of oscillating thing,so just multiply the whole thing by zero,why just zero? Because multiplying by any other number ain't going to help,it won't become a finite number.
∴ lim x → 0 x sin (1/2) = 0
The other way out is :
Now since you already have a x before sin (1/x) in the product form and the limiting value of x is 0... you can just directly say that zero will dominate the fight with the oscillating sin(1/x), that's because (→0) and the value of sin (1/x) oscillating from -1 to 1 ain't two opposite powers like some other values giving us the concept of indeterminate form,they are kind of friends,so the oscillation will sacrifice and let it's dear friend x dominate :)))
lim x → 0 x sin (1/x)
= (→0) sin (∞)
= (→0) (oscillating value from -1 to 1)
= 0
So,we are done with the intuitive ways, let's take the help of conceptual way now. So,the very popular way of doing such questions of the form f(x) (trigonometry function) is the sandwich/ squeeze theorem.
So,as we know the range of sin function is [-1,1] so any angle inside the sin will yield us some number between this range, using this we can write :
-1 ≤ sin (x) ≤ 1
Replace x by 1/x,
-1 ≤ sin (1/x) ≤ 1
Now in the middle try to make the function similar to the question you have been provided by performing mathematical operations, mostly you just have to use multiplication xD
So,we just have x missing out to make the inequality look similar to the question given,so just multiply by x,
-x ≤ x sin (1/x) ≤ x
Now either take the help of modulus or treat the three parts as three differentt limit question and evaluate the limit,
lim x → 0 - x ≤ lim x → 0 x sin (1/x) ≤ lim x → 0 x
(don't change the limiting value).
Now,we can easily calculate.
lim x → 0 - x = - 0 = 0
lim x → 0 x = 0
So, even the middle limit will be equal zero as per the sandwich theorem.