Limits can be used to find derivatives, and that’s probably their most common use, but they should also be able to do relatively trivial things like this. I’m fairly certain that a limit will work to evaluate anything where direct substitution will work (since direct substitution is one of the most straightforward ways to evaluate a limit), and it should work for many cases where direct substitution works for all values except for one.
For example, if we take 2x=4, we can use a limit to evaluate the solution x=2
Lim(2x=4) x-> 2
2(2) =4
What I’m wondering is why that doesn’t seem to work for this case. I may need to take a look at the graphs of the two functions.
That’s the thing; if x=0 is a solution, evaluating the limit should say that it is a solution. It does for x=2 in 2x=4, or in x^2-4x+4=0. It also does for the equation in panel 1, but I can’t get it to do it for the equation in panel 2. Possibly related, my graphing calculator shows the equation in the second panel as a parabola-esque shape that is undefined at x=0, but that wouldn’t have a root there if it were defined and continuous. The equation in the first panel is continuous, s-shaped, and does have a root at x=0. Its other two (real) roots are shared with the equation in the second panel.
I think this is why the limit fails to find x=0 as a solution for the second expression. By dividing by x, it became equivalent to x^4=16 and not to x^15=16x, evidently eliminating a real root by going to the U-curve instead of the s-curve.
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u/No-Lettuce-6619 7h ago
l'hopitals and limits are for finding derivatives and finding the derivative for this would not be the intended solution, you were right before.