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! =)
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!
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.
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.
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.
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.
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:
$\displaystyle sin(\frac{\pi}{6})=\frac{1}{2}$
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.
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:
$\displaystyle lim_{x\to 0} \frac{sin(x)}{x}=1$
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
$