# Limit question

1. Sep 16, 2011

### vrmuth

i am also stuck with this one
lim(x=>0) ( (x+9)^(1/2)-(x-9)^(1/2))/x , i wanna evaluate this algebraically,can anybody give me a clue

2. Sep 16, 2011

### JHamm

Re: Basic Limit Question

A couple of applications of L'Hopitals rule will do it :)

3. Sep 16, 2011

### vrmuth

Re: Basic Limit Question

hi jhamm thanks,could you please show how L'hopitals rule will do it , and cann't it be done algebraically?

4. Sep 16, 2011

### Hootenanny

Staff Emeritus
As you have written it, the limit does not exist. Have you copied the question correctly?

5. Sep 16, 2011

### SammyS

Staff Emeritus
The domain for the function $\displaystyle f(x)=\frac{\sqrt{x+9}-\sqrt{x-9}}{x}$ is [9, ∞) .

So, as Hootenanny wrote, there appears to be something wrong with the problem as you posted it.

6. Sep 17, 2011

### vrmuth

yes the limit does exist, thanks

7. Sep 17, 2011

### Harrisonized

Are you sure?

Let f(x) = g(x)/h(x).

Then:

lim f(x)
= lim [g(x)/h(x)]
= [lim g(x)] / [lim h(x)]

8. Sep 17, 2011

### vrmuth

did you see the function and that x tends to 0 ? :)

9. Sep 17, 2011

### Harrisonized

Let lim g(x) = a, where a∈ℂ (that is, a is a complex variable / it admits complex values). Then lim a/h(x) as h(x)→0 = ∞'​, where ∞'​ denotes the complex infinity (which, as its name suggests, doesn't exist on ℝ, the set of real numbers).

I'm telling you, the limit doesn't exist. Also fyi, L'hopital's rule doesn't work all the time.

10. Sep 17, 2011

### symbolipoint

vrmuth,
As described, for your given function, as x approaches 0, makes little sense because the domain cannot include any x value less than 9. You could try some algebraic tricks if you like, but you still have your originally given function. Try using a graphing calculator or a graphing program to display how the function looks, and check what you see "as x approaches 0".

11. Sep 18, 2011

### vrmuth

actually i 've written "does exist " instead of "doesn't exist" ,sorry,thanks for everybody

12. Sep 18, 2011

### Dickfore

13. Sep 18, 2011

### vrmuth

14. Oct 14, 2011

### vrmuth

can you please show me some example where L'hopital's rule won't work ?

15. Oct 14, 2011

### Harrisonized

lim x→∞ x/√(x2+1)

This limit actually came up in my electrostatics exam a few days ago.

16. Oct 14, 2011

### Dickfore

Last edited by a moderator: Apr 26, 2017
17. Oct 14, 2011

### Harrisonized

That's great. It's obviously 1. -_-

I'm just providing an example of l'Hopital's rule failing for the limit.

Last edited: Oct 14, 2011
18. Oct 14, 2011

### vrmuth

wow! its getting reciprocated each time we apply L'hopital rule , then what's the method to find such limits when l'hopital rule doesn't work ?

19. Oct 14, 2011

### Dickfore

Let $L \equiv \lim_{x \rightarrow \infty} \frac{x}{ \sqrt{x^2 + 1} }$. This is indeterminate form of the type $\frac{\infty}{\infty}$. Applying the L'Hospital's Rule with:
$$f(x) = x \Rightarrow f'(x) = 1$$
$$g(x) = \sqrt{x^2 + 1} \Rightarrow g'(x) = \frac{1}{2} (x^2 + 1)^{-\frac{1}{2}} 2 x = \frac{x}{\sqrt{x^2 + 1}}$$

Then:
$$\frac{f'(x)}{g'(x)} = \frac{\sqrt{x^2 + 1}}{x}$$

But, notice that this is the reciprocal of the original fraction! So, we have:
$$L = \frac{1}{L} \Rightarrow L^2 = 1 \Rightarrow L = \pm 1$$
The negative limit is impossible since both of the functions are positive. Thus, we are left with [itex]L = 1[/jtex].
So, L'Hospital's Rule does work in this case.

20. Oct 15, 2011

### VietDao29

There's one small error in your work: you assume that the limit does exist to apply L'Hopital's Rule, while the fact that this limit does exist, or not, is still unknown.

@vrmuth:

To solve these types of problem, we often divide both numerator, and denominator by x to the greatest power (in this problem is x).

$$\lim_{x \rightarrow +\infty} \frac{x}{\sqrt{x^2 + 1}}$$
$$=\lim_{x \rightarrow +\infty} \frac{\frac{x}{x}}{\frac{\sqrt{x^2 + 1}}{x}}$$
$$=\lim_{x \rightarrow +\infty} \frac{1}{\sqrt{\frac{x^2 + 1}{x^2}}} = ...$$

It should be easy to go from here. Let's see if you can get 1 as the answer. :)

Regards,