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:
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Rewrite the expression as ( \frac{5x + 1}{x + 2} ) since ( x^{-1} = \frac{1}{x} ).
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Apply l'Hôpital's Rule by taking the derivatives of the numerator and denominator with respect to ( x ).
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Differentiate the numerator: ( (5x + 1)' = 5 ).
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Differentiate the denominator: ( (x + 2)' = 1 ).
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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 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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