How do you find #lim (5+x^1)/(1+2x^1)# as #x>oo# using l'Hospital's Rule or otherwise?
# lim_( x rarr oo) (5+x^1)/(1+2x^1) =5#
having satisfied ourselves that our limit meets L'Hôpital's criteria to get
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To find ( \lim_{x \to \infty} \frac{5 + x^{1}}{1 + 2x^{1}} ) using l'Hôpital's Rule:

Rewrite the expression as ( \frac{5x + 1}{x + 2} ) since ( x^{1} = \frac{1}{x} ).

Apply l'Hôpital's Rule by taking the derivatives of the numerator and denominator with respect to ( x ).

Differentiate the numerator: ( (5x + 1)' = 5 ).

Differentiate the denominator: ( (x + 2)' = 1 ).

Evaluate the limit as ( x ) approaches infinity: ( \lim_{x \to \infty} \frac{5}{1} = 5 ).
Alternatively, you can directly evaluate the limit by observing that as ( x ) approaches infinity, the terms ( x^{1} ) become negligible compared to the constant terms, resulting in ( \frac{5}{1} = 5 ).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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