lim[(sin(x))/x]^(1/x^2) x趋近于0

来源：蜘蛛抓取(WebSpider) 时间：2017-12-27 14:12 标签： limsin2x x
Prove that limit (x^2)*[sin(1/x)]=0, x-&0? | eNotes
Prove that limit (x^2)*[sin(1/x)]=0, x-&0?
`lim_(x-&0) x^2 sin (1/x)=lim_(x-&0) x lim_(x-&0) sin(1/x)/(1/x)=1 lim_(x-&0) x=0`
To prove that lt x--&0 {x^2)[sin(1/x)] = 0.
Method 1: |sin theta|
& 1, for any theta.
So Lt x--& 0 x^2 {sin(1/x)}
should belong to
, x^2) as x--&0,
which is obviously zero .
Second method:
Consider a circle of radius
P on the circumference, centre O , making an angle theta with the initial line OX. Or angle XOP = theta.
be the perpendicular from P to OX. Then for any angle theta,
arc length (for the angle ) along the circumference.
r sin theta &
r theta. Or
sin theta & theta. Or
sin (1/x) & 1/x . Mutiplying
sin (1/x) & (1/x)x^2 = x.
Now as x--&0, x^2 sin(1/x) & x = 0.
For the beginning, we'll have to make the remark that we can't use the product law here (the limit of a product is the product of limits).
This thing happens because of the fact that lim sin(1/x) does not exist, when x tends to 0.
Though, -1 =& sin (1/x) =& 1
Now, if we'll multiply the inequality above, by x^2, because x^2 is positive, for any value of x, the inequality remains unchanged.
-x^2 =& (x^2)*sin (1/x) =& x^2
If we'll calculate lim x^2 = lim -x^2 = 0.
Now, we'll apply the Squeeze Theorem and we'll get :
lim -x^2=& lim (x^2)*sin (1/x) =& lim x^2
0=& lim (x^2)*sin (1/x) =&0
lim (x^2)*sin (1/x) = 0.
The evaluation of the limit is the one given by the enunciation.
The limit `lim_(x-&0) (x^2)*sin(1/x)` has to be determined.
If `y = 1/x` , as the value of x increases, the value of y decreases and conversely as x decreases, the value of y increases.
If `y = 1/x` and x tends to 0, y tends to `oo` .
In determining `lim_(x-&0) (x^2)*sin(1/x)` , we don't need to consider what the value of sin(1/x) becomes as x tends to 0, as the product of 0 and any number is 0. The square of x as x tends to 0 also tends to 0.
This gives:
`lim_(x-&0) (x^2)*sin(1/x)`
= `0^2*sin(1/0)`
The limit `lim_(x-&0) (x^2)*sin(1/x)` is equal to 0.
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x趋近于0时,x[sin(1/x^2)]的极限是多少?
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x→0lim x*sin(1/x^2)因为x趋于0,为无穷小量sin(1/x^2)为有界量无穷小量乘以有界量为无穷小量故,原极限=0有不懂欢迎追问
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a, b, c取何实数值才能使lim（x→0)（1／sin x－ax)∫x b（t^2／√1＋t^2)dt=c成立.
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求极限lim[sin（x-1）]/（x^2-1）x趋近于1
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lim【sin（x-1）】/（x^2-1）=lim【（x-1）/（x^2-1）】=lim【1/（x+1）】当x趋近于1时,该极限值为1/2注：sin（x-1）~（x-1）
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