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

Question

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

Evelyn Agard · Accepted Answer

We start by checking to make sure the limit is actually of the form #0/0# or #oo/oo#.

#L = 0^2/(sqrt(2(0) + 1) - 1) = 0/(1 - 1) = 0/0#

So we may indeed use l'hospital's.

The derivative of #sqrt(2x + 1)# is #2/(2sqrt(2x + 1)) = 1/sqrt(2x+ 1)#.

#L = lim_(x->0) (2x)/(1/sqrt(2x + 1)) #

#L = lim_(x->0) 2xsqrt(2x + 1)#

If we evaluate now, we get:

#L = 2(0)sqrt(2(0) + 1) = 0#

Looking at the graph, we realize that #y# does approach #0# as #x# approaches #0#.

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

Hopefully this helps!

Evelyn Agard · Answer

undefined 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:
$$ \lim_{x \to 0} \frac{x^2}{\sqrt{2x+1} - 1} = \frac{0}{1} = 0 $$

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

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