How do you determine the limit of #5^x/(3^x+2^x)# as x approaches infinity?

Answer 1

Use l'Hopital's Rule and some algebra.

In the original form, l'Hopital's rule doesn't help, but note that:

#5^x/(3^x+2^x)=(5^x/2^x)/(3^x/2^x+1)=(5/2)^x/((3/2)^x+1)#.
The limit as #xrarroo# is of the form #oo/oo#.

L'Hopital tells us to consider the ratio of the derivatives:

#((5/2)^xln(5/2))/((3/2)^xln(3/2)#.
Note that this is simply #(5/2)^x/(3/2)^x# times the positive constant #ln(5/2)/ln(3/2)#.
Furthermore, #(5/2)^x/(3/2)^x=(5/3)^x#.
So, as #xrarroo#, #5^x/(3^x+2^x)# behaves like #((5/2)^xln(5/2))/((3/2)^xln(3/2)# which in turn behaves like #k(5/3)^x# with #k>0#. Which increases without bound (goes to #oo#) as #xrarroo#.
#lim_(xrarroo)5^x/(3^x+2^x)=oo#.
Bonus: #lim_(xrarr-oo)(5/2)^x/((3/2)^x+1)=0/(0+1)# so #lim_(xrarr-oo)5^x/(3^x+2^x)=0#
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Answer 2

To determine the limit of 5^x/(3^x+2^x) as x approaches infinity, we can use the concept of limits and properties of exponential functions.

First, let's rewrite the expression as (5/3)^x / (1 + (2/3)^x).

As x approaches infinity, (5/3)^x will tend to infinity since the base (5/3) is greater than 1.

Similarly, (2/3)^x will tend to 0 as x approaches infinity since the base (2/3) is between 0 and 1.

Therefore, the limit of (5/3)^x / (1 + (2/3)^x) as x approaches infinity is infinity.

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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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