...lim(x-->0)((sqrt(x+5) - sqrt(5))/x), an indeterminate case of 0/0 after direct substitution, (rewrite the denominator x as (x+5) -5) = (sqrt(x+5) - sqrt(5))(sqrt(x+5) + sqrt(5)); I treated (x+5) - 5 as a difference of squares) --> the original limit after the final cancelling of top and bottom common factors becomes: lim(x-->0)(1/(sqrt(x+5) + sqrt(5)) = 1/2sqrt(5) = sqrt(5)/10. An alternative way of solving the same limit, Jan-W
1. To determine in the first place whether it's defined or undefined. 2. For a limit approaching zero with a denominator that would equal zero, this doesn't mean there is no limit. It means we have to do more work to fijd it a different way. A limit is not the behavior AT a point that csn be undefined; it's defined as the behavior NEAR/APPROACHING that value, so the limit exists and acts the same regardless of whether that point is non-continuous/undefined/has a denom of 0.
sir, how would root 10 equal 2root5. Wouldn't root20 equal 2root5? maybe I'm missing something here, but wouldn't you break up 20 into 5 and 4 and take the root of 4 which would be 2 , so then it would be 2root5?
he substituted in 0 for x so it became 1/sqrt((0)+5)+sqrt(5) then 0+5 = 5 so it became 1/sqrt(5) + sqrt(5). then combine like terms (x+x = 2x) it became 1/2sqrt(5)
Justin Lee In the rest of the world and school, it's standard form to always rationalize denominators. You would clear the fraction like any other. Multiply a single root by itself, or a sum/difference by its conjugate. To keep the expression equivalent, your factor must = 1, so put the same number on top and bottom of a fraction. For example in trigonometry, if you have tan(30°), it's 1/2 over √3/2 which simplifies to 1/√3. The is improper form, so we need to clear the radical. To keep from changing the value, multiply by √3/√3. Now you have √3/3!
This concept gives you the slope at any point, and that's necessary to move father in math, with precalculus, differentials, integrals, physics, some trig, etc.
the best teacher ever I understand every single video you upload much appreciate it
1:48 "anyone have any questions on what i have done"
I... I have a question
.....
Great example, thank you!
How did you cancel x by x , the x was adding in numerator
Thank you very much
3:59 sigh of depression. Hahaha nice video.
when you realize you're a mathematician but you still can't get her number. 3:59
...lim(x-->0)((sqrt(x+5) - sqrt(5))/x), an indeterminate case of 0/0 after direct substitution, (rewrite the denominator x as (x+5) -5) = (sqrt(x+5) - sqrt(5))(sqrt(x+5) + sqrt(5)); I treated (x+5) - 5 as a difference of squares) --> the original limit after the final cancelling of top and bottom common factors becomes: lim(x-->0)(1/(sqrt(x+5) + sqrt(5)) = 1/2sqrt(5) = sqrt(5)/10. An alternative way of solving the same limit, Jan-W
The limit is continuous as x approaches 0 and is equal to 1÷2√5 when x=0 but becomes discontinuous or undefined when x
Thankyou sir!
how did the denominator become 2 square root of 5?
Think of root 5 like an x, (1x + 1x = 2x). Now replace x with root 5. You'll get 2 times root 5.
Thanks!
What would happen if you didn't change x by 5???
I just tried that, following the question in the thumbnail which had that error, the answer was not defined
You are amazing
Why solve for it if its undefined?
1. To determine in the first place whether it's defined or undefined.
2. For a limit approaching zero with a denominator that would equal zero, this doesn't mean there is no limit. It means we have to do more work to fijd it a different way.
A limit is not the behavior AT a point that csn be undefined; it's defined as the behavior NEAR/APPROACHING that value, so the limit exists and acts the same regardless of whether that point is non-continuous/undefined/has a denom of 0.
sir, how would root 10 equal 2root5. Wouldn't root20 equal 2root5? maybe I'm missing something here, but wouldn't you break up 20 into 5 and 4 and take the root of 4 which would be 2 , so then it would be 2root5?
root 5 + root 5 isn't equal to root 10. Think of root 5 like an x, (1x + 1x = 2x). Now replace x with root 5. You'll get 2 times root 5.
Emmm, teacher, I don't under stand why you put 2raiz 5
that is the square root conjugate, as you notice once I multiply it through I no longer have a radical
@@brianmclogan hello still donf get it
he substituted in 0 for x so it became 1/sqrt((0)+5)+sqrt(5) then 0+5 = 5 so it became 1/sqrt(5) + sqrt(5). then combine like terms (x+x = 2x) it became 1/2sqrt(5)
the point of manipulating the whole thing is to get it to the point where substituting in the value you want doesnt cause 0/0 or undefined etc.
hello sir 😊
I don't understand the question was radical x but you later replaced with 5
he made a mistake writing it down. radical 5 was the actual problem
@@Naijiri. l see thank you
What must i do if the root is in a the denominator?! Will it be the same way ?!?!!
You can rationalise it if you want to, but it’s not necessary usually if it’s on the AP exam unless it’s mentioned.
Justin Lee In the rest of the world and school, it's standard form to always rationalize denominators.
You would clear the fraction like any other. Multiply a single root by itself, or a sum/difference by its conjugate. To keep the expression equivalent, your factor must = 1, so put the same number on top and bottom of a fraction.
For example in trigonometry, if you have tan(30°), it's 1/2 over √3/2 which simplifies to 1/√3. The is improper form, so we need to clear the radical. To keep from changing the value, multiply by √3/√3. Now you have √3/3!
Where did that 2 radical sign 5 came from
The x goes to 0, so the only thing left in the denominator is the sqrt 5 + sqrt 5, which is equal to 2 times sqrt 5
We say it 2 root 5 in india😂
The limit doesn't exist for this problem because of the fact that you get a non zero number over zero when you use direct substitution right away
Check the graph, I did write down the problem wrong and fixed it later in the video
The limit DNE, because from the left of zero sqrt(x) does not exists.
Nice
lim x → 0 root(1+x) - root(1-x) / x = ?
Sorry just able to focus on what I am currently teaching
try mathway.com
Thank you Brian ANS is 1
How did the final answer end up to be 2 root 5?🤷🏽♂️
Can you solve a doubt
Limit of x approach to zero 4xsquer +5xque +7xsquer +5x Divided by 5xsqure +8xto the power 7+x
Please reply me sir
can pls anyone tell me where we can use limit t->0?
what do you mean?
where to use limit "t" aproaches to zero??"sir"
t is just a variable like x in this problem, it does not matter what variable you use
sir my ques 's ans is that ......we use t approaches to zero when there very small change in tym .....moreover thnx srr😉😊
This concept gives you the slope at any point, and that's necessary to move father in math, with precalculus, differentials, integrals, physics, some trig, etc.
Sorry. My mistake
no worries, I should changed the thumbnail!
Ok ok I got it
i got the square root of 5 over 10 . dont know why
That’s only if you rationalize the denominator. The answer is still the same.
Esse caboclo não me inspira segurança.
No le sabes
No su mathematica es muy mal.
من الرفش ماله
نسه يكلب الاشاير مال عامل منسب هههه
Waste
okay
How did the final answer end up to be 2 root 5?🤷🏽♂️
How did the final answer end up to be 2 root 5?🤷🏽♂️