lim (x-cos x)/(2x+10) as abs|x| approaches to infinity ?

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embizze's profile pic

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

Find `lim_(x->oo)(x-cosx)/(2x+10)`

(1) Write as `lim_(x->oo)(x-cosx)/(2x+10)=lim_(x->oo)[(x)/(2x+10)-(cosx)/(2x+10)]`

(2) The limit of a sum is the sum of the limits:

`lim_(x->oo)[x/(2x+10)-(cosx)/(2x+10)]=lim_(x->oo)x/(2x+10)-lim_(x->oo)(cosx)/(2x+10)`

(3) `lim_(x->oo)x/(2x+10)=1/2` . (Multiply numerator and denominator by `1/x` to get `lim_(x->oo)1/(2+10/x)` . Then since `lim(x->oo)10/x=0` we get the result)

(4) `lim_(x->oo)(cosx)/(2x+10)=0` . The numerator oscillates from 1 to -1, and the denominator increases without bound.Consider `-2/(2x+10)<(cosx)/(2x+10)<2/(2x+10)` . But `lim_(x->oo)(-2)/(2x+10)=lim_(x->oo)2/(2x+10)=0`

(5) Then `lim_(x->oo)(x-cosx)/(2x+10)=1/2+0=1/2`

(6) By similar arguments `lim_(x->-oo)x/(2x+10)=1/2` and `lim_(x->-oo)(cosx)/(2x+10)=0`

Therefore, `lim_(|x|->oo)(x-cosx)/(2x+10)=1/2`

Here is a graph for checking purposes: x:scale from -40 to 40, y-scale from -1 to 1

minhthien94's profile pic

minhthien94 | Student, Undergraduate | (Level 1) eNoter

Posted on

We have

     -1 <= cos x <= 1

So

     (x+1)/(2x+10) >= (x-cosx)/(2x+10) >= (x-1)/(2x+10)

But

     as x approaches to infinity, both (x+1)/(2x+10) and

     (x-1)/(2x+10) approach to 1/2

Therefore

     as x approaches to infinity, (x-cosx)/(2x+10) approaches to 1/2.

[Squeeze Theorem]

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