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Answer 1 · 2017-09-02T12:05:26

1

So we know that on the real numbers, #sin# and #cos# are bounded. This means that

#-1<=cos(x)<=1# and #-1<=sin(x)<=1# #AA# #x in RR#

So we can manipulate the limit to take advantage of this. We start with

#lim_(xrarroo) (x^2 + sin(x))/(x^2+cos(x))#

Now divide top and bottom by #x^2#, the limit becomes:

#lim_(xrarroo) (1 + (sin(x))/(x^2))/(1+(cos(x))/(x^2))#

As x tends to infinity #sin(x)# and #cos(x)# remain bounded however #x^2 rarr oo#, hence

#lim_(xrarroo) sin(x)/x^2 = 0# and #lim_(xrarroo) cos(x)/x^2 = 0#

So we have that

#lim_(xrarroo) (1 + (sin(x))/(x^2))/(1+(cos(x))/(x^2)) = 1/1 = 1#