# How do you find #lim x^2/(sqrt(2x+1)-1)# as #x->0# using l'Hospital's Rule?

So we may indeed use l'hospital's.

If we evaluate now, we get:

graph{x^2/(sqrt(2x+ 1) - 1) [-4.93, 4.934, -2.465, 2.465]}

Hopefully this helps!

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To find ( \lim_{x \to 0} \frac{x^2}{\sqrt{2x+1} - 1} ) using L'Hôpital's Rule, first notice that the denominator approaches (0) as (x) approaches (0). Therefore, both the numerator and the denominator approach (0) as (x) approaches (0).

Apply L'Hôpital's Rule by differentiating both the numerator and the denominator with respect to (x) separately.

Differentiate the numerator: [ \lim_{x \to 0} \frac{d}{dx}(x^2) = \lim_{x \to 0} 2x = 0 ]

Differentiate the denominator: [ \lim_{x \to 0} \frac{d}{dx}(\sqrt{2x+1} - 1) = \lim_{x \to 0} \frac{1}{2\sqrt{2x+1}} \cdot 2 = \lim_{x \to 0} \frac{1}{\sqrt{2x+1}} = \frac{1}{\sqrt{1}} = 1 ]

Now, take the ratio of the derivatives: [ \lim_{x \to 0} \frac{x^2}{\sqrt{2x+1} - 1} = \frac{0}{1} = 0 ]

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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.

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