let u = tan x. Then, limit u-> \infty => lim x -> pi/2. Lim u-> \infty u/ \sqrt(1+u^2) => lim (x-> \pi/2 ) tan x / \sqrt( 1+ tan^2 x) = lim (x-> pi/2) sin x = 1
In case you're wondering what would happen if you try to use L'Hopital's rule for this kind of limit, the thing that would happen is you will still get the inf/inf indeterminate form no matter how many times you differentiate the numerator and denominator because whenever you differentiate a square root function, the part with the whole square root will always repeat itself. For example in this case, the derivative of sqrt(u^2 + 1) is u/sqrt(u^2 + 1), and when simplified after differentiating will be the reciprocal (that is only if you are trying to evaluate this specific limit). Trying to evaluate that expression as x goes to infinity will give you inf/inf again. The main point is that the derivative of any square root function will always repeat itself along with chain rule in its expression if necessary. So, whenever you're trying to evaluate this kind of limit, do so with the method shown in the video. Sorry for the long explanation.
@@chadmongoose I think it is a very good pointer actually because most people just smack with LH whenever they see any inf/inf or 0/0 and forgetting about all the other methods we have to solve limits algebraically. And this is especially important if LH rule results in a DNE limit because if that is the case you cannot conclude anything about the original limit, forcing you to use other methods and approaches. (BPRP and also micheal penn have done some examples of these)
You could use squeeze theorem. Let L= u/sqrt(u²+1) u/sqrt(u²) > L > u/sqrt(u²+2u+1) Thus 1 > L > u/(u+1) On the limit, we have 1>= lim L >= 1 Thus lim L = 1
In this instance, sqrt(u²) = u because we're dealing with the limit as u goes to infinity, meaning that all values of u > 0. But in general, sqrt(u²) = |u| which is where the resolution for the limit of negative infinity will come from: |u| = -u when u
I substituted u=tan(x), then the whole expression simplifies to sin(x), which is just sin(tan-1(u)). Now as u→infinity, tan-1(u)→π/2 and sin(π/2) is just 1.
Here's how I did it with L'H's rule. Let L equal the original limit: L = limit as u to inf of sqrt(u^2 + 1)/u dN/du = 1 dD/du = 1/(2*sqrt(u^2 + 1)) * 2*u dD/du = u/sqrt(u^2 + 1) Construct (dN/du)/(dD/du): sqrt(u^2 + 1)/u Observe that this is a reciprocal of the original limit. Thus: L = 1/L The only numbers that equal their own reciprocal, are +1 and -1. Since both the top and bottom are positive for the domain of the limit in question, the limit must therefore equal 1.
3:25 in my first calc 1 test i was going to get a sweet 10/10, but the professor didn't accept my resolution of solving a limit to infinity with something similar to the trick you used because it breaks some of the rules of root limits. This method surely works for multiple choice questions, but if your calculus course is a bit more rigorous you are better off going with the first method
Yea it’s one of those things where the math all works out but sometimes you need to explain why you did what you did otherwise the professor/teacher will get mad
If u->infinity, then an some point it must be positive. Because its derivative is its reciprocal and the function is always positive when u is positive, the function is strictly increasing when u is positive. But also it is bounded above by 1 because √(u²+1)>u. Since it is strictly increasing and bounded above, it must converge. NOW L'Hospital's rule tells us its limit is 1 because the limit is positive and equal to it's own reciprocal.
That is essentially what we are doing by “comparing the dominating part” because it is essentially saying that all the lower powers of the variable become so small that they are essentially insignificant as it tends to +/-inf
I have always been confused by the limit of a function if only one side exists. The simplest example is the limit as sqrt(x) as x approaches 0. Is it 0 or DNE? Maybe prove it using the definition.
I was taught ti use the O(x) method, by checking the behavior at info the dominant exponentials. So I would have calculated the limit as X->oo of u/sqrtu^2) -> u/u=1
Well, strictly speaking, you can, it just doesn't get you anywhere useful. The rule does still apply, the original limit is equal to the limit of the derivative of the numerator divided by the derivative of the denominator, it just happens that, in this case, the derivates don't actually simplify the original expression...
Think of how you solve Looper Integration By Parts problems, like integral sin(x)*e^x dx. You probably ended up with something like this: sin(x)*e^x - cos(x)*e^x - integral sin(x)*e^x dx Observe how the original integral appears in it. Assign I to equal the original integral, and equate the whole expression to I: I = sin(x)*e^x - cos(x)*e^x - I And solving for I, algebraically, allows us to conclude the solution is: 1/2*e^x * [sin(x) - cos(x)] + C You can do a similar thing with this limit. Let L equal the original limit: L = limit as x to inf: x/sqrt(x^2 + 1) Take derivatives of top and bottom, as we do with L'H's rule: L = lim as x to inf: 1/(2*x/(2*sqrt(x^2 + 1)) After simplifications we get: L = lim as x to inf: sqrt(x^2 + 1)/x Take the reciprocal of both sides: 1/L = 1/(lim as x to inf: sqrt(x^2 + 1)/x) And simplify: 1/L = lim as x to inf: x/sqrt(x^2 + 1) Observe that we now have the original limit, thus: 1/L = L There are only two numbers that equal their own reciprocal, which are -1 and +1. Since sqrt(x^2 + 1) can never be negative for real x's, and since we're interested in a positive value of x, we can conclude that the only valid solution is L=1.
let u = tan x. Then, limit u-> \infty => lim x -> pi/2.
Lim u-> \infty u/ \sqrt(1+u^2) => lim (x-> \pi/2 ) tan x / \sqrt( 1+ tan^2 x) = lim (x-> pi/2) sin x = 1
This is a very nice way!
In case you're wondering what would happen if you try to use L'Hopital's rule for this kind of limit, the thing that would happen is you will still get the inf/inf indeterminate form no matter how many times you differentiate the numerator and denominator because whenever you differentiate a square root function, the part with the whole square root will always repeat itself. For example in this case, the derivative of sqrt(u^2 + 1) is u/sqrt(u^2 + 1), and when simplified after differentiating will be the reciprocal (that is only if you are trying to evaluate this specific limit). Trying to evaluate that expression as x goes to infinity will give you inf/inf again. The main point is that the derivative of any square root function will always repeat itself along with chain rule in its expression if necessary. So, whenever you're trying to evaluate this kind of limit, do so with the method shown in the video. Sorry for the long explanation.
ok thanks
@@chadmongoose I think it is a very good pointer actually because most people just smack with LH whenever they see any inf/inf or 0/0 and forgetting about all the other methods we have to solve limits algebraically.
And this is especially important if LH rule results in a DNE limit because if that is the case you cannot conclude anything about the original limit, forcing you to use other methods and approaches. (BPRP and also micheal penn have done some examples of these)
@@Ninja20704 yeah thats what i did so i just thanked him
@@chadmongoose oh sorry but I originally meant to reply to the OP, but good that you think the same too.
You could use squeeze theorem.
Let L= u/sqrt(u²+1)
u/sqrt(u²) > L > u/sqrt(u²+2u+1)
Thus
1 > L > u/(u+1)
On the limit, we have
1>= lim L >= 1
Thus lim L = 1
In this instance, sqrt(u²) = u because we're dealing with the limit as u goes to infinity, meaning that all values of u > 0. But in general, sqrt(u²) = |u| which is where the resolution for the limit of negative infinity will come from: |u| = -u when u
Good point! Very important warning for this method!
I substituted u=tan(x), then the whole expression simplifies to sin(x), which is just sin(tan-1(u)). Now as u→infinity, tan-1(u)→π/2 and sin(π/2) is just 1.
Same here
Haha the L'hospital's rule is pretty funny. Just an infinitely spinning equation.
Here's how I did it with L'H's rule.
Let L equal the original limit:
L = limit as u to inf of sqrt(u^2 + 1)/u
dN/du = 1
dD/du = 1/(2*sqrt(u^2 + 1)) * 2*u
dD/du = u/sqrt(u^2 + 1)
Construct (dN/du)/(dD/du):
sqrt(u^2 + 1)/u
Observe that this is a reciprocal of the original limit. Thus:
L = 1/L
The only numbers that equal their own reciprocal, are +1 and -1. Since both the top and bottom are positive for the domain of the limit in question, the limit must therefore equal 1.
3:25 in my first calc 1 test i was going to get a sweet 10/10, but the professor didn't accept my resolution of solving a limit to infinity with something similar to the trick you used because it breaks some of the rules of root limits. This method surely works for multiple choice questions, but if your calculus course is a bit more rigorous you are better off going with the first method
Yea it’s one of those things where the math all works out but sometimes you need to explain why you did what you did otherwise the professor/teacher will get mad
If u->infinity, then an some point it must be positive. Because its derivative is its reciprocal and the function is always positive when u is positive, the function is strictly increasing when u is positive. But also it is bounded above by 1 because √(u²+1)>u. Since it is strictly increasing and bounded above, it must converge. NOW L'Hospital's rule tells us its limit is 1 because the limit is positive and equal to it's own reciprocal.
just keep the highest power for both numerator and denominator.
I love that L'H rule flips the expression. I'm guessing because the answer is 1 that it's like flipping 1/1 over and over.
Hyperbolic sine also correct? u:=sinh(x). With cosh^2(x) - sinh^2(x) = 1, follows u / sqrt(u^2 +1) = sinh(x) / cosh(x) = tanh(x) -> 1 for x to inf.
Can we take analytic approach?
u^2+C ~ u^2 as u -> infinity
lim u/sqrt(u^2+1) ~ lim u/sqrt(u^2) = lim u / |u| = 1
That is essentially what we are doing by “comparing the dominating part” because it is essentially saying that all the lower powers of the variable become so small that they are essentially insignificant as it tends to +/-inf
This is not rigorous and can easily lead you to a mistake
Hmm, this is how I teach it... I guess I have to give it another thought
Nvm this is mentioned at the end of the video ;)
I have always been confused by the limit of a function if only one side exists. The simplest example is the limit as sqrt(x) as x approaches 0. Is it 0 or DNE? Maybe prove it using the definition.
I would say that would be a bad question in the first place. It only makes sense to ask the limit of sqrt(x) as x ->0+
I was taught ti use the O(x) method, by checking the behavior at info the dominant exponentials. So I would have calculated the limit as X->oo of u/sqrtu^2) -> u/u=1
This can be quite tricky if differences are involved instead of ratios, so you need to be careful
Never said I would always use automatically the O(x) method
now prove it given epsilon>0 chooose M>0, sps u>M, check...
Can I do this:
lim(u→∞) u/√(u²+1) = lim(u→∞) 1/√(1+1/u²) = 1
If 1/infinity is 0 , so 0 times infinity is 1
I substitute u = tan theta
Yeah 😂
If you apply l hopital rule it will be a forever loop.
But sir why can't I use l hopital rule for this one? 🤔
Well, strictly speaking, you can, it just doesn't get you anywhere useful. The rule does still apply, the original limit is equal to the limit of the derivative of the numerator divided by the derivative of the denominator, it just happens that, in this case, the derivates don't actually simplify the original expression...
Think of how you solve Looper Integration By Parts problems, like integral sin(x)*e^x dx. You probably ended up with something like this:
sin(x)*e^x - cos(x)*e^x - integral sin(x)*e^x dx
Observe how the original integral appears in it. Assign I to equal the original integral, and equate the whole expression to I:
I = sin(x)*e^x - cos(x)*e^x - I
And solving for I, algebraically, allows us to conclude the solution is:
1/2*e^x * [sin(x) - cos(x)] + C
You can do a similar thing with this limit. Let L equal the original limit:
L = limit as x to inf: x/sqrt(x^2 + 1)
Take derivatives of top and bottom, as we do with L'H's rule:
L = lim as x to inf: 1/(2*x/(2*sqrt(x^2 + 1))
After simplifications we get:
L = lim as x to inf: sqrt(x^2 + 1)/x
Take the reciprocal of both sides:
1/L = 1/(lim as x to inf: sqrt(x^2 + 1)/x)
And simplify:
1/L = lim as x to inf: x/sqrt(x^2 + 1)
Observe that we now have the original limit, thus:
1/L = L
There are only two numbers that equal their own reciprocal, which are -1 and +1. Since sqrt(x^2 + 1) can never be negative for real x's, and since we're interested in a positive value of x, we can conclude that the only valid solution is L=1.
@@carultch thanks man 🙏🏻