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

[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?

 

[VietDao29]

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? :)

 

[marlon]

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

 

[Johnnycab]

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?

 

[HallsofIvy]

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.

 

[Johnnycab]

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

 

[nothing123]

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).

 

[arildno]

What are BEDMAS rules??

 

[nothing123]

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