How do you find the limit #lim (3^x-2^x+1)/(4*3^x-2^x-1)# as #x->oo#?
# lim_(x rarr oo) (3^x-2^x+1)/(4*3^x-2^x-1) =1/4#
We seek:
We can manipulate the limits as follows:
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To find the limit as x approaches infinity, we can analyze the behavior of the numerator and denominator separately.
As x approaches infinity, the term 3^x grows much faster than 2^x and 1. Therefore, the term 2^x+1 becomes negligible compared to 3^x.
Similarly, in the denominator, the term 43^x grows much faster than 2^x and 1. Thus, the term 2^x-1 becomes negligible compared to 43^x.
By simplifying the expression, we can rewrite it as (3^x/3^x) / (4*3^x/3^x).
Simplifying further, we get 1 / 4.
Therefore, the limit of (3^x-2^x+1)/(4*3^x-2^x-1) as x approaches infinity is 1/4.
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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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- How do you calculate this limit without using l’Hospital’s rule: lim √(x^2 +1) -1 / √(x^2+16) - 4 as x → 0 ?

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