continuity itself means that the limit of a function at a point is equal to its value at that point, performing the switcheroo is applying that principle in reverse order: the value at a point is equal to the limit of the function at that point (because = goes both ways)
@@nobodyspecial7895 when there's a discontinuity, the graph "teleports" at one point Take f(x) = x-1 for x < 5, x+1 for x ≥ 5 The limit on the left side of 5 is equal to 4, and on the right side it's equal to 6, meaning the limit at 5 doesn't exist You can also look up Dirichlet function, you have f(1)=1, but the limit at 1 doesn't exist (easily proven with epsilon-delta as the values never get closer together)
for some reason i did the following: lim {ln(1-cosx) - ln(x^2)} as x->0 lim [ln{(1-cosx)/x^2}] as x->0 let L = lim [ln{(1-cosx)/x^2}] as x->0 e^L = lim (1-cosx)/x^2 as x->0 now to focus on the RHS lim {(1-cosx)/x^2} as x->0 we get 0/0, so we use L'Hopital's rule after which, we get, lim (sinx/2x) as x->0 =(1/2)*{lim (sinx/x) as x->0} the limit in the latter portion equals 1 =1/2 so e^L=1/2 L=ln(1/2) = -ln(2)
Since lim_(x->0) sin x / x cannot be solved by L' Hospital's rule, lim_(x->0) [(1 - cos x)/x^2] also cannot be solved by L' Hospital's rule since the derivative of cos x also depends on lim_(x->0) sin x / x. The correct way should be lim_(x->0) [(1 - cos x)/x^2] = lim_(x->0) {[2sin^2 (x/2)]/x^2} = 1/2* lim_(x->0) [sin (x/2) / (x/2)]^2 = 1/2* { lim_(x->0) [sin (x/2) / (x/2)] }^2 = 1/2 * 1^2 = 1/2
@@ramennoodle9918 I don't think using l'Hôpital's is ever required, in fact it's usually recommended against unless you're learning it for the first time. So in this case, it turns out to be much simpler to use Taylor expansions.
No, your post is wrong. 1) You never mentioned the x going to *what limit.* 2) You are missing needed grouping symbols in the numerator: [1 - cos(x)]/x^2.
Calculus 1, how do you evaluate these limits (not 0/0 form) Reddit r/homeworkhelp
th-cam.com/video/tCzh__jf-u4/w-d-xo.html
Sir I ahev some problem
Can you see it pls
I'm aware of the inside-out changeability due to ln being a continuous function, but I never learned why that's the case.
continuity itself means that the limit of a function at a point is equal to its value at that point, performing the switcheroo is applying that principle in reverse order: the value at a point is equal to the limit of the function at that point (because = goes both ways)
@@konradybcio Thanks. Clears things up.
@@konradybcio When is it not the case that a point on a graph would be different to the limit of that same x value?
@@nobodyspecial7895 when there's a discontinuity, the graph "teleports" at one point
Take f(x) = x-1 for x < 5, x+1 for x ≥ 5
The limit on the left side of 5 is equal to 4, and on the right side it's equal to 6, meaning the limit at 5 doesn't exist
You can also look up Dirichlet function, you have f(1)=1, but the limit at 1 doesn't exist (easily proven with epsilon-delta as the values never get closer together)
@@konradybcio Thanks for this answer, I'll have to look into it more.
for some reason i did the following:
lim {ln(1-cosx) - ln(x^2)} as x->0
lim [ln{(1-cosx)/x^2}] as x->0
let L = lim [ln{(1-cosx)/x^2}] as x->0
e^L = lim (1-cosx)/x^2 as x->0
now to focus on the RHS
lim {(1-cosx)/x^2} as x->0
we get 0/0, so we use L'Hopital's rule
after which, we get,
lim (sinx/2x) as x->0
=(1/2)*{lim (sinx/x) as x->0}
the limit in the latter portion equals 1
=1/2
so e^L=1/2
L=ln(1/2) = -ln(2)
Since lim_(x->0) sin x / x cannot be solved by L' Hospital's rule, lim_(x->0) [(1 - cos x)/x^2] also cannot be solved by L' Hospital's rule since the derivative of cos x also depends on lim_(x->0) sin x / x.
The correct way should be
lim_(x->0) [(1 - cos x)/x^2]
= lim_(x->0) {[2sin^2 (x/2)]/x^2}
= 1/2* lim_(x->0) [sin (x/2) / (x/2)]^2
= 1/2* { lim_(x->0) [sin (x/2) / (x/2)] }^2
= 1/2 * 1^2
= 1/2
Why can't you apply L'Hospital to lim x-> 0 sin (x) / x? It's of the form 0/0, and L'Hospital gives cos x / 1 and cos(0)/1 equals 1....
1-cosx ~ x²/2 (Taylor series) therefore we have
ln(x²/2)-ln(x²)=ln½, done
this is if using L’Hôpital wasn’t required?
@@ramennoodle9918 I don't think using l'Hôpital's is ever required, in fact it's usually recommended against unless you're learning it for the first time. So in this case, it turns out to be much simpler to use Taylor expansions.
Please put Turkish subtitles
I just did. Please check and let me know if it works!
@@bprpcalculusbasicsIt works, thank you.☺️
The limit of 1-cos (x)/x^2 is a famous limit (=1/2)
No, your post is wrong. 1) You never mentioned the x going to *what limit.* 2) You are missing needed grouping symbols in the numerator: [1 - cos(x)]/x^2.