How do you find $lim_(x->∞)(x+2sinx-1)/(x+3cosx+1)$ ?

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Answer 1 · 2018-01-18T15:01:19

$lim_(xrarr+oo)(x+2sinx-1)/(x+3cosx+1)=1$

$lim_(xrarr+oo)(x+2sinx-1)/(x+3cosx+1)$

$x->+oo$
$x>0$

$=lim_(xrarr+oo)(1+2sinx/x-1/x)/(1+3cosx/x+1/x)=1$

because

$|sinx/x|<=1/|x|$
so

$-1/|x|<=sinx/x<=1/|x|$

Using the sandwich/squeeze theorem we get:

$lim_(xrarr+oo)-1/|x|=0=lim_(xrarr+oo)1/|x|$

Therefore ,

$lim_(xrarr+oo)sinx/x=0$

$|cosx/x|<=1/|x|$

$-1/|x|<=cosx/x<=1/|x|$

$lim_(xrarr+oo)-1/|x|=0=lim_(xrarr+oo)1/|x|$

Therefore,

$lim_(xrarr+oo)cosx/x=0$

How do you find lim_(x->∞)(x+2sinx-1)/(x+3cosx+1)lim_(x->∞)(x+2sinx-1)/(x+3cosx+1)?