I really love how you're doing college level math while approaching everything like you're speaking in front of a bunch of 6th graders. Not the kind of patience and calmness everyone has.
You are awesome sir! I'm now very intersted in math after your videos! I can't speak your language, but try my best to understand everything you explain
To all of you who suggested 0.25 as the limit (like I did): that's indeed the limit when x goes to infinity - but not the limit when x goes to negative infinity, as the problem had it 😉 like my math teacher said: "take time to read the problem text carefully". Which I clearly didn't ...
have my Calculus exam tomorrow and your videos are so consice and easy to understand. And the way youn explain somehow calms my nerves😂.Your videos have come in clutch
In the square root it's probably possible to factorise by x^7. This way we can get it to become the sqrt of x^7(4x^7+1) which is plus infinity ,and by adding -2x^7 the result will still be the same mayby.
Hi, I have an idea.When x approche -Infinity, √(4x^14+x^7) is approximately =√4x^14.Since √4x^14 is positive, so √4x^14=2 (absolute value x )^7, which is positive infinity, and if we -2x^7, it equals to4(absolute value x)^7, which is also positive infinity. I think this way is more simple
I would have solved it like this: let X>0 then I would have divided the radicand by x^14. Then the limit would have been |/(4+1/x^2) - 27x^7 => sqrt(4+0) - 27^-inf > 2+inf => +inf
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I really love how you're doing college level math while approaching everything like you're speaking in front of a bunch of 6th graders. Not the kind of patience and calmness everyone has.
I studied math at uni and this video is amazing 👍. The calm, clear and concise explanation is nice.
I did it by factoring.
I don't know if I'm right but here is how I did it:
You are awesome sir! I'm now very intersted in math after your videos! I can't speak your language, but try my best to understand everything you explain
To all of you who suggested 0.25 as the limit (like I did): that's indeed the limit when x goes to infinity - but not the limit when x goes to negative infinity, as the problem had it 😉 like my math teacher said: "take time to read the problem text carefully". Which I clearly didn't ...
despite being a pre-calc student im really good at guessing these limits and its mostly looking at which x is 'bigger' even tho there both inf
I know how to solve these limits but you always find some ways i don't know.
We can solve this without any calculations too.
have my Calculus exam tomorrow and your videos are so consice and easy to understand. And the way youn explain somehow calms my nerves😂.Your videos have come in clutch
In the square root it's probably possible to factorise by x^7. This way we can get it to become the sqrt of x^7(4x^7+1) which is plus infinity ,and by adding -2x^7 the result will still be the same mayby.
Very nice problem and brilliant explications. Thanks a lot
شكرا استاذ تعلمت منك الكثير
Very nice thank you.
You can argue that the 14th power is a higher order infinity. Or that the end behavior of the polynomial would tail off at infinity on the right side.
Hi, I have an idea.When x approche -Infinity, √(4x^14+x^7) is approximately =√4x^14.Since √4x^14 is positive, so √4x^14=2 (absolute value x )^7, which is positive infinity, and if we -2x^7, it equals to4(absolute value x)^7, which is also positive infinity. I think this way is more simple
To infinity, and beyond . . .
I would have solved it like this: let X>0 then I would have divided the radicand by x^14. Then the limit would have been |/(4+1/x^2) - 27x^7 => sqrt(4+0) - 27^-inf > 2+inf => +inf
I always prefer to change limit to -inf to inf , specially when there is square root in the problem
What background is this?