Evaluating lim(x=>0) (x+9)^(1/2)-(x-9)^(1/2)/x

[vrmuth]

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

 

[JHamm]

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

 

[vrmuth]

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

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

 

[Hootenanny]

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

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

 

[SammyS]

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

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.

 

[vrmuth]

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.

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

yes the limit does exist, thanks

 

[Harrisonized]

yes the limit does exist, thanks

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)]

 

[vrmuth]

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)]

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

 

[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.

 

[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".

 

[vrmuth]

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.

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".

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

 

[Dickfore]

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

yes the limit does exist, thanks

Ok, then, the answer is:
http://www.wolframalpha.com/input/?i=Limit[%28x%2B9%29^%281%2F2%29-%28x-9%29^%281%2F2%29%29%2Fx%2Cx-%3E0]

 

[vrmuth]

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.

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

Ok, then, the answer is:
http://www.wolframalpha.com/input/?i=Limit[%28x%2B9%29^%281%2F2%29-%28x-9%29^%281%2F2%29%29%2Fx%2Cx-%3E0]

let's end this thread

 

[vrmuth]

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

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

 

[Harrisonized]

lim x→∞ x/√(x²+1)

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

 

[Dickfore]

lim x→∞ x/√(x²+1)

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

http://www.wolframalpha.com/input/?i=Limit[x/Sqrt[x^2+++1],+x+->+Infinity]"

 

[Harrisonized]

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

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

 

[vrmuth]

lim x→∞ x/√(x²+1)

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

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 ?

 

[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.

 

[VietDao29]

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.

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,

 

[Dickfore]

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.

Well, you can look at the use of l'Hospital's Rule as a heuristic guide. It does give a correct solution if we make the assumption that a finite limit exists. For those that require a rigorous proof, one can then go ahead and construct an epsilon-delta proof by using these identities:

$$\left\vert \frac{x}{\sqrt{x^2 + 1}} - 1 \right\vert = \left\vert \frac{x - \sqrt{x^2 + 1}}{\sqrt{x^2 + 1}} \right\vert = \left\vert \frac{x - \sqrt{x^2 + 1}}{\sqrt{x^2 + 1}} \cdot \frac{x + \sqrt{x^2 + 1}}{x + \sqrt{x^2 + 1}} \right\vert = \left\vert \frac{x^2 - x^2 - 1}{\sqrt{x^2 + 1} \left( x + \sqrt{x^2 + 1} \right)} \right\vert$$

$$\frac{1}{\sqrt{x^2 + 1} \left \vert x + \sqrt{x^2 + 1} \right \vert} = \frac{1}{\sqrt{x^2 + 1} \left( x + \sqrt{x^2 + 1} \right)}$$

Next, we notice that $\sqrt{x^2 + 1}$ and $x + \sqrt{x^2 + 1}$ are both monotonically increasing, and the following inequality:
$$\sqrt{x^2 + 1} > |x|, \forall x$$
holds (the graph of the hyperbola lies above its asymptotes). For $x > M > 0$, we have the following series of inequalities:
$$\sqrt{x^2 + 1} \left( x + \sqrt{x^2 + 1} \right) > \sqrt{M^2 + 1} \left( M + \sqrt{M^2 + 1} \right) > M (M + M) > 2 M^2$$

For any given $\epsilon > 0$, we can find such an $M > M_{0} > 0$, so that:
$$2 M^2_0 = \frac{1}{\epsilon}$$
$$M_0 = \frac{1}{\sqrt{2 \epsilon}}$$

$$\left( \forall \epsilon > 0 \right) \left(\exists M_0(\epsilon) > 0 \right) \left( \forall x > M > M_{0} \right) \left\vert \frac{x}{\sqrt{x^2 + 1}} - 1 \right \vert \equiv \frac{1}{\sqrt{x^2 + 1} \left( x + \sqrt{x^2 + 1} \right) } < \epsilon$$

Thus, we had constructed a proof that:
$$\lim_{x \rightarrow \infty}{\frac{x}{\sqrt{x^2 + 1}}} = 1$$

 

[vrmuth]

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

Regards,

yes, i can get , thanks :)