Question #11ebf

1 Answer
Sep 2, 2017

1

Explanation:

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