Multiply through $1/x$, and also hence, together outlined by the other response, the difficulty reduces to present what $frac cos x x $ ideologies to as $x o infty $.
Hint: due to the fact that $| cos y | leq 1 $ for every $y$, we have actually
$$ frac-1x leq frac cos x x leq frac1x $$
Now, use the to express Rule.
You are watching: Limit as x approaches infinity of cosx
$$lim_x o infty 2x-cos(x) over 3x+cos(x) $$$$=lim_x o infty 2 - cos(x)/x over 3 +cos(x)/x$$$$(1) quad =2 - lim_x o infty left( cos(x) over x ight) over 3 +lim_x o infty left( cos(x) over x ight) $$
Cosine deserve to only take it on worths from 1 to -1 for this reason it has finite bound. In the limit, it"s separated by an unlimited number therefore the limit of...
$$lim_x o infty cos(x) over x=0$$
Thus, (1) amounts to $ 2 over 3$ and therefore,
$$lim_x o infty 2x-cos(x) over 3x+cos(x) = 2 over 3$$
edited Jun 19 "15 at 14:29
reply Jun 19 "15 at 14:24
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by using long division$$frac2x-cos x3x+cos x=frac23-frac5/3cos x3x+cos x$$
so the limit is$$lim_x ightarrow infty frac2x-cos x3x+cos x=lim_x ightarrow infty frac23-lim_x ightarrow infty frac5/3*cos x3x+cos x=frac23-0=frac23$$
edited Jun 19 "15 at 14:32
answered Jun 19 "15 in ~ 14:23
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If we factor an $x$ from your rational function, then us have
$$frac2x-cos(x)3x+cos(x) = fracxleft(2-fraccos(x)x ight)xleft(3+fraccos(x)x ight) = frac2-fraccos(x)x3+fraccos(x)x$$
You may recall the identification $lim_x o inftyleft(fraccos(x)x ight) = 0$, i m sorry can quickly be evidenced using L"Hopital"s dominance for indeterminates that the type $fracinftyinfty$.
Evaluating the limit, us have
$$lim_x oinftyleft(frac2x-cos(x)3x+cos(x) ight) = lim_x oinftyleft(frac2-fraccos(x)x3+fraccos(x)x ight) = frac2-lim_x oinftyleft(fraccos(x)x ight)3+lim_x oinftyleft(fraccos(x)x ight) = frac23$$
which is the equipment to the problem.
answer Jun 19 "15 in ~ 14:36
J. DunivinJ. Dunivin
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