How do you find #lim (x^3+4x+8)/(2x^32)# as #x>1^+# using l'Hospital's Rule or otherwise?
The limit does not exist (it is infinite).
The limit does not exist (it is infinite).
graph{(x^3+4x+4)/(2x^32) [10, 10, 25, 25]}
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To find lim (x^3+4x+8)/(2x^32) as x approaches 1 from the right using l'Hôpital's Rule, differentiate the numerator and the denominator separately with respect to x, then evaluate the limit of the resulting expressions as x approaches 1 from the right.

Differentiate the numerator: f'(x) = 3x^2 + 4

Differentiate the denominator: g'(x) = 6x^2

Substitute x = 1 into the derivatives: f'(1) = 3(1)^2 + 4 = 7 g'(1) = 6(1)^2 = 6

Evaluate the limit of the ratio of the derivatives: lim (x>1^+) f'(x)/g'(x) = lim (x>1^+) (3x^2 + 4)/(6x^2) = 7/6
So, lim (x^3+4x+8)/(2x^32) as x approaches 1 from the right is 7/6.
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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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