I am supposed to find this limit without L'Hopital's rule. I have tried multiplying the numerator and the denominator with (1/(1-x))+1, and I've looked for ways to factorize, without getting anywere. Would appreciate any help! =)

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- Jul 5th 2010, 12:28 PMgralla55limit problem
I am supposed to find this limit without L'Hopital's rule. I have tried multiplying the numerator and the denominator with (1/(1-x))+1, and I've looked for ways to factorize, without getting anywere. Would appreciate any help! =)

- Jul 5th 2010, 12:39 PMAlso sprach Zarathustra
$\displaystyle lim \frac{\frac{1}{x+1} -1}{x}=lim \frac{-1}{x+1}$

so when $\displaystyle x \to 0$ the above equals to $\displaystyle -1$ - Jul 5th 2010, 01:04 PMgralla55
Thank you very much! Would you mind explaining exactly how you simplified the limit? Did you just muliply the numerator and the denominator with (x+1)? I would never have spotted that, thanks again!

- Jul 5th 2010, 03:45 PMAlso sprach ZarathustraOf course...
$\displaystyle lim \frac{\frac{1}{x+1} -1}{x}=lim \frac{\frac{1-1 \cdot (x+1)}{x+1}}{x}=lim \frac{\frac{1-x-1}{x+1}}{x}=lim \frac{\frac{-x}{x+1}}{x}=lim \frac{-x}{x(x+1)}=lim\frac{-1}{x+1}$

- Jul 5th 2010, 04:05 PMAlso sprach Zarathustra
And something more...

I notice that you have $\displaystyle 11^2$ posts already and yet you haven't try write math with Latex, I strongly recommend you to try it!

You can start learning this here, on this site there is a forum about Latex...

...but the best way I think is by double-clicking on the code, you can try it with my post above!

Anyway... good-luck!(Cool) - Jul 5th 2010, 07:38 PMtheodds
Another quick way to see this is to note that if $\displaystyle f(x) = \frac{1}{1 + x}$ then

$\displaystyle \displaystyle

\lim_{x \to 0} \frac{\frac{1}{1 + x} - 1}{x} = f'(0)

$

by definition. Given the way the limit was set up, this is probably what was expected. - Jul 6th 2010, 02:43 PMgralla55
Thanks alot! I tried using Latex for this next problem, but I kept getting errors. I will have to start using it on easier expressions first I think... Anyway, in this problem, is it possible for me to go from step 2 to step 3 like I did here? In that case, the rest of the problem is easy.

- Jul 6th 2010, 02:49 PMtheodds
You can do this using the same trick I mentioned. If $\displaystyle f(x) = \sin x$ then

$\displaystyle \displaystyle

\lim_{h \to 0} \frac{f \left(\frac{\pi}{6} + h \right) - 1/2}{h} = f'\left(\frac{\pi}{6}\right) = \cos \frac{\pi}{6}.

$

To get at the same thing you are suggesting here, note that you can rewrite the second expression as

$\displaystyle \displaystyle

\frac{1}{2} \frac{(\cos x - 1)}{x} + \cos \left(\frac{\pi}{6}\right) \frac{\sin x}{x}.

$

Then take the limit, using the usual results for this types of things. - Jul 6th 2010, 02:56 PMgralla55
That will indeed work! I've never seen anyone using the defintion of the derivative in solving limit problems like that, haha. But if I do it like I did, then it will also simplify to 1 times cos (pi/6) which is what you got. I'm just not sure if I can rewrite it like that. It would be nice to know for future problems.

- Jul 6th 2010, 02:58 PMAlso sprach Zarathustra
No! This is wrong...

But you can solve this problem by definition of derivative:

$\displaystyle f'(x_0)=lim_{\Delta{x}\to 0} \frac{f(x+\Delta{x})-f(x_0)}{\Delta{x}}$

Hint:

$\displaystyle f(x)$ in your case is: $\displaystyle f(x)=sin(x)}$

And you know that:

$\displaystyle sin(\frac{\pi}{6})=\frac{1}{2}$ - Jul 6th 2010, 03:00 PMtheodds
Yeah, see the edit I made. It works it out explicitly. What you can do is, if $\displaystyle \lim f(x) = A$ and $\displaystyle \lim g(x) = B$ then the following hold:

$\displaystyle \displaystyle

\lim(f(x) + g(x)) = A + B \qquad \lim f(x) g(x) = AB \qquad \lim \frac{f(x)}{g(x)} = \frac{A}{B}

$

provided, in the last case, that $\displaystyle \lim g(x) \ne 0$. You can't do what you did though, as a general rule. - Jul 6th 2010, 03:13 PMgralla55
Thank you both! Very helpful! Using the last limit properties, can I solve the problem like this?

- Jul 6th 2010, 03:19 PMAlso sprach Zarathustra
ANOTHER SOLUTION:

Recall the trigonometric identity:

$\displaystyle sin(x)-sin(y)=2cos(\frac{x+y}{2})sin(\frac{x-y}{2})$

Now, as we know: $\displaystyle \frac{1}{2}=sin(\frac{\pi}{6})$

Therefore:

$\displaystyle sin(\frac{\pi}{6}+x)+\frac{1}{2}=sin(\frac{\pi}{6} +x)+sin(\frac{\pi}{6})=2cos(\frac{\pi}{6}+\frac{x} {2})sin(\frac{x}{2})$

Now, you can continue this using:

$\displaystyle lim_{x\to 0} \frac{sin(x)}{x}=1$ - Jul 6th 2010, 03:23 PMAlso sprach Zarathustra
- Jul 6th 2010, 03:30 PMtheodds
You can't move the limit taking into the denominator, because $\displaystyle \lim_{x \to 0} x = 0$. The rule for moving limits around doesn't allow for this. The result would be undefined to begin with.

The limit results you want to use (taken as given) are:

$\displaystyle \displaystyle

\lim_{x \to 0} \frac{\sin x}{x} = 1, \qquad \lim_{x \to 0} \frac{\cos x - 1}{x} = 0

$