# 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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1 year 2021-08-24T01:33:03+00:00 1 Answers 0 views 0

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.