Find the limit as x approaches infinity of #y=arccos((1+x^2)/(1+2x^2))#?

Answer 1
There is a law of limits that deals with composite functions. Essentially, if we have two functions #f# and #g#:
#lim_(x->a) f(g(x)) = f(lim_(x->a) g(x))#
In this case, #f(g(x))# would be #arccos(g(x))#, and #g(x)# would be #(1+x^2)/(1+2x^2)#.
So, to find the limit of the entire thing as #x# approaches infinity, we can find the limit of the inner function as #x# approaches infinity, and then evaluate the outer function with this value:
#lim_(x->infty) arccos((1+x^2)/(1+2x^2)) = arccos(lim_(x->infty) (1+x^2)/(1+2x^2))#
It should be easy to see that #lim_(x->infty) (1+x^2)/(1+2x^2)# is equal to #1/2#.

So, we have:

#lim_(x->infty) arccos((1+x^2)/(1+2x^2)) = arccos(1/2)#
The arccosine of #1/2# is equal to #pi/3#:
#lim_(x->infty) arccos((1+x^2)/(1+2x^2)) = pi/3#
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Answer 2

The limit as x approaches infinity of y=arccos((1+x^2)/(1+2x^2)) is π/2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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