MHB Solve Limits & Algebra Problem: Get $\dfrac{{v_0}^2}{2gsin\alpha }$

[DanielBW]

Hello guys ! I need your help with the next problem:

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"Show that the equation:

$\dfrac{mv_0}{k}-\dfrac{m^2g}{k^2}sin\alpha \cdot ln\left[ 1+\dfrac{kv_0}{mgsin\alpha } \right]$

is equivalent to:

$\dfrac{{v_0}^2}{2gsin\alpha }$

when k tends to 0 ($k\rightarrow 0)$"
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Solving i get:

$\lim\limits_{k\rightarrow 0}\dfrac{mv_0}{k}-\lim\limits_{k\rightarrow 0}\dfrac{m^2g}{k^2}sin\alpha \cdot ln\left[ 1+\dfrac{kv_0}{mgsin\alpha } \right]$

where the second limit has the 0/0 situation, so i can use L'Hopital rule on it:

$\lim\limits_{k\rightarrow 0}\dfrac{mv_0}{k}-\lim\limits_{k\rightarrow 0}\left[ \dfrac{m^2g}{2k}sin\alpha \cdot \left(\dfrac{mgsin\alpha }{mgsin\alpha + kv_0} \right) \left(\dfrac{v_0}{mgsin\alpha }\right) \right]$

Simplifying and re-ordening i get:

$\lim\limits_{k\rightarrow 0}\dfrac{mv_0}{k} \cdot \lim\limits_{k\rightarrow 0} \left[1-\dfrac{mgsin\alpha }{mgsin\alpha +kv_0}\right]$

$\lim\limits_{k\rightarrow 0}\left[\dfrac{1}{k} \cdot \dfrac{mk{v_0}^2}{mgsin\alpha +kv_0}\right]$

$\lim\limits_{k\rightarrow 0}\left[\dfrac{m{v_0}^2}{mgsin\alpha +kv_0} \right]$

Therefore:

$\dfrac{{v_0}^2}{gsin\alpha }$

I can't find the 1/2 to show the equality... what am i doing wrong? Hope you guys can help me, i'd appreciate it !

 

[Nono713]

Your derivation is really messy and hard to follow, due to all the different $m$, $v_0$, $\sin\alpha$ constants floating everywhere. I believe you either made a simple error when calculating your derivatives, or "broke the rules" by doing limit arithmetic on indeterminate forms without realizing it. I see at least an $\infty - \infty$ in your solution which seems to have been overlooked (at the beginning where you split up the limit into two at the minus sign - you can't just do that as both tend to infinity) but it is unclear whether it's the cause of the problem.

For reference, here is a 'tidied up' solution which you might be able to follow to find out where exactly you made your mistake. It is in a spoiler box to give you the option to ignore it.

Spoiler

Let us substitute $c = v_0 / (mg\sin\alpha)$ to reduce clutter, which is fine since those are presumably just constants independent of $k$ (I chose this particular substitution so the logarithm would come out as simple as possible). We want to find the value of the following limit (please verify for yourself it is equivalent to your original limit by substituting $c$ back in):

$$L = \lim_{k \to 0} \left [ \frac{mv_0}{k} - \frac{m v_0}{c k^2} \ln(1+ kc) \right ]$$

Note that we can already pull the $m v_0$ factor out, leaving us with:

$$L = m v_0 \lim_{k \to 0} \left [ \frac{1}{k} - \frac{1}{c k^2} \ln(1+ kc) \right ]$$

And thus write $L' m v_0 = L$ to get rid of the last constants. All the annoying extra symbols have now disappeared under a veil of mathematical abstraction, leaving us free to focus on the limit without distraction. In any case, we are left with:

$$L' = \lim_{k \to 0} \left [ \frac{1}{k} - \frac{\ln(1+ kc)}{c k^2} \right ]$$

At this point note that we have an indeterminate form already: the first term $1/k$ tends to infinity, while the other one (less obviously) tends to infinity as well (try to prove this). Thus we have an indeterminate form of the type $\infty - \infty$ and l'Hôpital's rule applies after converting it to a suitable indeterminate quotient:

$$L' = \lim_{k \to 0} \left [ \frac{c k}{c k^2} - \frac{\ln(1+ kc)}{c k^2} \right ] = \lim_{k \to 0} \left [ \frac{ck - \ln(1 + kc)}{c k^2} \right ]$$

At that point we clearly have an indeterminate form of the type $0/0$. Differentiating numerator and denominator, we get:

$$L' = \lim_{k \to 0} \left [ \frac{c - \frac{c}{1 + kc}}{2 c k} \right ]$$

Note this is where your missing factor of two pops up. We simplify by cancelling out the $c$ and multiplying out the fraction:

$$L' = \lim_{k \to 0} \left [ \frac{1 - \frac{1}{1 + kc}}{2 k} \right ] = \lim_{k \to 0} \left [ \frac{(1 + kc) - 1}{2 k (1 + kc)} \right ] = \lim_{k \to 0} \left [ \frac{kc}{2 k (1 + kc)} \right ]$$

This gives us the opportunity to cancel out the $k$ term which was preventing us from evaluating the limit:

$$L' = \lim_{k \to 0} \left [ \frac{c}{2 (1 + kc)} \right ] = \lim_{k \to 0} \left [ \frac{c}{2 + 2kc} \right ] = \frac{c}{2}$$

Substituting everything back, we get:

$$L = m v_0 \frac{c}{2} = m v_0 \frac{v_0}{2 mg\sin\alpha} = \frac{v_0^2}{2 g\sin\alpha}$$

As expected.

 

[DanielBW]

Hello Bacterius! Thank you very much for help me, as you said, I've been mistaken in the "split up" step. After correcting that mistake I was able to get your solution "$\dfrac{c}{2}$" in the form $\dfrac{v_0}{2mgsin\alpha }$ to finally get the expected solution.

Again thank you a lot ! It really helps the substitution to reduce the clutter.