Can the Undefined Limit of a Velocity Equation be Solved for as k Approaches 0?

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The discussion focuses on evaluating the limit of a velocity equation as k approaches 0, specifically lim(k-->0) [ (-mg)/k + v*e^(kt/m) + (mg)/k*e^(kt/m)]. The expected outcome is to derive the velocity of an object at any time t while neglecting air resistance, which should yield v - gt. The equation originates from the differential equation dv/dt - k/m * v = g, leading to a solution involving integrating factors. Despite attempts to manipulate the equation using logarithms and exponent rules, the user struggles to eliminate terms that complicate the limit evaluation. Ultimately, setting k to 0 in the original differential equation suggests a different solution, highlighting the complexity of the limit process.
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Undefined Limits as k--> 0

Homework Statement



lim(k-->0) [ (-mg)/k + v*e^(kt/m) + (mg)/k*e^(kt/m)]

,the end result of this limit is ultimately supposed to be v -gt (or the velocity of an object at any time t neglecting air resistance).



Homework Equations



This equation comes from the differential equation dv/dt - k/m *v =g

,then using integrating factors (the equation itself is a linear ODE) I found:

v = (-mg)/k + ce^(kt/m)

,where c is found by solving for the initial condition v(0)=v0 where

v0 = (-mg)/k + c(1)

--> c = v0 + mg/k

The Attempt at a Solution



I've spent literally a few hours pouring over this, frustrated as hell that I couldn't solve a simple limit!

I tried first taking the natural log of the whole thing, then I tried using all of the rules of exponents separating the e^(x)s. Then I tried working with the differential equation for a while but ultimately was never able to find a way to fully end up with a defined answer - ie. I was unable to completely eliminate x^(-1) or ln(x) prior to taking the limit as x-->0.
 
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Rewrite the solution as ( I hope I haven't made a mistake)

v \left( t \right) = \exp \left( \frac{k}{m} t \right) v_0 +gm \frac{\exp \left( \frac{k}{m} t \right) - 1}{k},

and then take the limit k \rightarrow 0.

Note that if you set k = 0 in the original differential equation, the solution is v = v_0 + gt, not v = v_0 - gt.
 
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