# Evaluate Limit with L'Hôpital's Rule: x => 0 of sqrt(x^2+x) - x

• nothing123
In summary, based on the information provided, it does not appear that the limit for this expression will ever be reached.
nothing123
evaluate the limit as x=>0 of sqrt(x^2+x) - x

so i multiplied by the conjugate to get:

x^2 + x - x^2 / sqrt(x^2 +x) + x

which simplifies to:

x / sqrt (x^2 + x) + x = infinity/infinity

taking LH, you get:

1 / 1/2(x^2+x)^(-1/2) * (2x+1) + 1 = infinity/infinity again

looking ahead, it doesn't look like LH is going to help

any suggestions?

nothing123 said:
evaluate the limit as x=>0 of sqrt(x^2+x) - x
Are you sure it's not infinity?
Suppose you mean:
$$\lim_{x \rightarrow \infty} \sqrt{x ^ 2 + x} - x$$
If $$x \rightarrow - \infty$$, you will have: $$\infty + \infty = \infty$$, the limit does not exist for that case, so here, we are going to find the limit of that expression as x tends to positive infinity.
$$\lim_{x \rightarrow + \infty} \sqrt{x ^ 2 + x} - x$$
so i multiplied by the conjugate to get:

x^2 + x - x^2 / sqrt(x^2 +x) + x

which simplifies to:

x / sqrt (x^2 + x) + x = infinity/infinity
This is good.
Do you need to use L'Hospital rule? If not, you can consider dividing both numerator, and denominator by x to get:
$$\lim_{x \rightarrow + \infty} \sqrt{x ^ 2 + x} - x = \lim_{x \rightarrow + \infty} \frac{x}{\sqrt{x ^ 2 + x} + x}= \lim_{x \rightarrow + \infty} \frac{\frac{x}{x}}{\frac{\sqrt{x ^ 2 + x} + x}{x}}$$
$$= \lim_{x \rightarrow + \infty} \frac{1}{\sqrt{1 + \frac{1}{x}} + 1}$$, (since x is positive ($$x \rightarrow + \infty$$), so $$\sqrt{x ^ 2} = x$$)
Can you go from here? :)

nothing123 said:
evaluate the limit as x=>0 of sqrt(x^2+x) - x

Well, in this case you don't need to multiply by the conjugate because if you just evaluate this function with x being 0 you get 0. This function is continuous in x = 0 so you can just calculate the limit of f(x) by calculating f(0).

marlon

sorry to cut into this post but I am working atm on a question similar to this, i haven't even seen LH rule in my book so that may be why I am going to ask this question

how did you turn all those x's into 1's and 1/x in the numerator?

Standard step you should have learned as the first thing in limits: if you are taking limits as x goes to infinity, divide numerator and denominator by the highest power of x. That way all the "x"s become "1/x"s and instead of going to infinity, they go to 0.

HallsofIvy said:
Standard step you should have learned as the first thing in limits: if you are taking limits as x goes to infinity, divide numerator and denominator by the highest power of x. That way all the "x"s become "1/x"s and instead of going to infinity, they go to 0.

yep that makes sense, thanks

VietDao, based on your calculations, I get 1/2 as an answer. However, I don't know how your supposed to divide sqrt (x^2 + x) by x. Doesn't this violate BEDMAS rules b/c I thought sqrt(x^2+x) does not equal sqrt(x^2) + sqrt(x).

Last edited:
What are BEDMAS rules??

sorry nvm, i forgot that sqrt(a) divide sqrt(b) = sqrt(a/b). thanks for everyones help.

## 1. What is L'Hôpital's Rule?

L'Hôpital's Rule is a mathematical technique used to evaluate the limit of a function when the limit approaches a specific value, such as zero or infinity. It involves taking the derivative of both the numerator and denominator of the original function and then evaluating the limit again.

## 2. How is L'Hôpital's Rule used to evaluate limits?

L'Hôpital's Rule is used by taking the derivative of both the numerator and denominator of a function, and then evaluating the limit again. This process can be repeated until a definite value is obtained or until the limit is indeterminate.

## 3. What is the limit of sqrt(x^2+x) - x as x approaches 0?

The limit of sqrt(x^2+x) - x as x approaches 0 is 0. This can be evaluated using L'Hôpital's Rule, by taking the derivative of both the numerator and denominator, which results in 1/2x + 1. When x is plugged in as 0, the result is 0.

## 4. Can L'Hôpital's Rule be used to evaluate limits at other values?

Yes, L'Hôpital's Rule can be used to evaluate limits at other values besides 0. It can be used for limits approaching infinity or any other specific value.

## 5. Are there any limitations to using L'Hôpital's Rule?

Yes, there are limitations to using L'Hôpital's Rule. It can only be used for certain types of limits, such as indeterminate forms like 0/0 or ∞/∞. It also requires taking derivatives, so it may not be applicable for functions that are not differentiable.

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