can't we apply L'hopital's rule for all the limits questions ?
I know that it can be used only if it is an undetermined foam.But if it is a determined form,will there be a question to find the limit of?we can easily figure it out
can't we apply L'hopital's rule for all the limits questions ?
I know that it can be used only if it is an undetermined foam.But if it is a determined form,will there be a question to find the limit of?we can easily figure it out
In short, no you can't apply L'Hospital's Rule for every limit question. As you know, you can only apply L'Hospital's Rule to the indeterminate forms $\displaystyle \displaystyle \frac{0}{0}$ or $\displaystyle \displaystyle \frac{\infty}{\infty}$ (though you can usually transform other indeterminate forms to one of these forms), but not even every indeterminate form can have L'Hospital's Rule applied to it.
The perfect example is $\displaystyle \displaystyle \lim_{x \to 0}\frac{\sin{x}}{x}$.
The reason being because to use L'Hospital's Rule, you need to evaluate the derivatives of the numerator and the denominator, but evaluating the derivative of $\displaystyle \displaystyle \sin{x}$ involves evaluating $\displaystyle \displaystyle \lim_{\Delta x \to 0}\frac{\sin{\Delta x}}{\Delta x}$ - the exact same limit!
Dear Prove It,
Unless I am mistaken L'Hospital's rule can be applied if the limit of the derivative $\displaystyle \left(\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}\right)$ exists. So, for your example it does exist and L'Hospital's rule can be used.
$\displaystyle \displaystyle \lim_{x \to 0}\frac{\sin{x}}{x}=\displaystyle\lim_{x\to 0}\frac{\cos x}{1}=1$
I think Prove It's point is that the reasoning becomes circular. In order to take the derivative of sin(x), you need to evaluate the very limit in question. If you already know that the derivative of sin is cos, without having needed this limit to compute that derivative, then fine. I think most books show what this limit is before showing that sin'(x) = cos(x). So it's a logical problem, not a computational problem.
Ah...I understand it now. Thanks for the clarification. I am sure Prove It must have meant that. But think about it, if they had proved L'Hospital's rule first, it could be easily deduced that, $\displaystyle \displaystyle \lim_{x \to 0}\frac{\sin{x}}{x}=1$ I think this would be better because then there would be no need for a geometric construction (as I have seen in most books). Moreover the proof would be more general, because we can include complex numbers for x as well.
It does not matter what order "L'Hospital's Rule" and "the derivative of $\displaystyle \displaystyle \sin{x}$" were found in. In order to use L'Hospital's Rule for $\displaystyle \displaystyle \lim_{x \to 0}\frac{\sin{(x)}}{x}$, you need to be able to find the derivative of $\displaystyle \displaystyle \sin{(x)}$. Let's try this...
$\displaystyle \displaystyle \begin{align*} \frac{d}{dx}[\sin{(x)}] &= \lim_{\Delta x \to 0}\frac{\sin{(x + \Delta x)} - \sin{(x)}}{\Delta x} \\ &= \lim_{\Delta x \to 0}\frac{\sin{(x)}\cos{(\Delta x)} + \sin{(\Delta x)}\cos{(x)} - \sin{(x)}}{x} \\ &= \sin{x}\lim_{\Delta x \to 0}\frac{\cos{(\Delta x)} - 1}{\Delta x} + \cos{(x)}\lim_{\Delta x \to 0}\frac{\sin{(\Delta x)}}{\Delta x} \end{align*}$
Notice that you have to evaluate $\displaystyle \displaystyle \lim_{\Delta x \to 0}\frac{\sin{(\Delta x)}}{\Delta x}$, which is the EXACT same limit as $\displaystyle \displaystyle \lim_{x \to 0}\frac{\sin{(x)}}{x}$. How could you possibly argue that using L'Hospital's Rule is appropriate in this case, because evaluating the derivative of $\displaystyle \displaystyle \sin{(x)}$ implies you have already found the very limit you are trying to find? It's entirely circular reasoning.
Here's a trivial example.
$\displaystyle \lim_{x\to\infty} \frac{e^{-x}}{e^{-x}}$
Both numerator and denominator, separately go to 0 but it you try to use L'Hopitals rule you just keep getting
$\displaystyle \frac{e^{-x}}{e^{-x}}$
so that L'Hopital's rule does not work.
Of course, the limit is 1.
Here's another one (although not as trivial as Halls ex.) that sends you in a circular pattern
$\displaystyle \lim_{x \to \infty} \dfrac{x}{\sqrt{x^2+1}}$
Point is - when L'Hopital's method doesn't work, sometimes you have to be more creative.