What happens if you try to use L’Hopital’s Rule to find the limit? lim x/√x^2 +6 x→[infinity] Required:

Question

What happens if you try to use L’Hopital’s Rule to find the limit?

lim x/√x^2 +6
x→[infinity]

Required:
a. You cannot apply L’Hopital’s Rule because the function is not differentiable.
b. You cannot apply L’Hopital’s Rule because the numerator equals zero for some value x = a.
c. You cannot apply L’Hopital’s Rule because the function is not continuous.
d. You cannot apply L’Hopital’s Rule because the denominator equals zero for some value x = a.
e. Repeated applications of L’Hopital Rule result in the original limit or the limit of the reciprocal of the function.

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Diễm Thu 5 months 2021-08-24T01:33:03+00:00 1 Answers 0 views 0

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    2021-08-24T01:34:08+00:00

    Answer:

    e. Repeated applications of L’Hopital Rule result in the original limit or the limit of the reciprocal of the function.

    Step-by-step explanation:

    \lim_{x \to \infty} \frac{x}{\sqrt{x^{2} + 6} } =  ∞/√(∞² + 6)= ∞/∞

    Using L’Hopital’s Rule,

    \lim_{x \to \infty} \frac{x}{\sqrt{x^{2} + 6} } = \lim_{x \to \infty} \frac{\frac{dx}{dx} }{\frac{d\sqrt{x^{2} + 6}}{dx}  } =  \lim_{x \to \infty} \frac{\sqrt{x^{2} + 6}}{x} = √(∞² + 6)/ ∞ = ∞/∞

    Applying L’Hopital’s rule again, we have

    \lim_{x \to \infty} \frac{\sqrt{x^{2} + 6}}{x} = \lim_{x \to \infty} \frac{\frac{\sqrt{x^{2} + 6}}{dx} }{\frac{dx}{dx}} = \lim_{x \to \infty} \frac{x}{\sqrt{x^{2} + 6}}  } = ∞/√(∞² + 6)= ∞/∞

    Applying L’Hopital’s rule again, we have

    \lim_{x \to \infty} \frac{x}{\sqrt{x^{2} + 6} } = \lim_{x \to \infty} \frac{\frac{dx}{dx} }{\frac{d\sqrt{x^{2} + 6}}{dx}  } =  \lim_{x \to \infty} \frac{\sqrt{x^{2} + 6}}{x} = √(∞² + 6)/ ∞ = ∞/∞

    Applying L’Hopital’s rule again, we have

    \lim_{x \to \infty} \frac{\sqrt{x^{2} + 6}}{x} = \lim_{x \to \infty} \frac{\frac{\sqrt{x^{2} + 6}}{dx} }{\frac{dx}{dx}} = \lim_{x \to \infty} \frac{x}{\sqrt{x^{2} + 6}}  } = ∞/√(∞² + 6)= ∞/∞

    So, we see that repeated applications of L’Hopital Rule result in the original limit or the limit of the reciprocal of the function.

    So, e is the answer.

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