limit (5^x+4^x)^1/x as x goes to inf 

Another excellent explanation! Your forte is your pace. I.E. you do NOT rush & race through the explanation as a lot of professors do.😀

Mr PM, you are one of the best teachers I have ever seen. The way you keep the student hooked up all the way up to the final solution is just magic. And I have noticed some pedagogic tricks you have and I find them quite ingenious! I also like your way of not using too many intermediate steps.

Solution: ˣ√(5ˣ + 4ˣ) = ˣ√((1.25 * 4)ˣ + 4ˣ) = ˣ√(1.25ˣ * 4ˣ + 4ˣ) = ˣ√(4ˣ * (1.25ˣ + 1)) = ˣ√4ˣ * ˣ√(1.25ˣ + 1) = 4 * ˣ√(1.25ˣ + 1) In the Limit the "+1" become negligible and we end up with: = 4 * ˣ√1.25ˣ = 4 * 1.25 = 5

Dirty maths: For x>1: 5^x < 5^x + 4^x < 2*(5^x) We can take all to the (1/x) power since 1/x is stricture decreasing towards infinity Then we get: 5 < (5^x + 4^x)^(1/x) < 5 * (2^(1/x)) Taking x to infinity squeezes our limit between to 5. Yay!

What I really like is I can watch this with the sound off and it is still just as clear -> the steps are all correct - great stuff. We all knew it must be 5 intuitively comparing 5^x and 4^x at infinity - but intuition is not math!! You must derive it the right way. Great to go back in time to my undergraduate days and see a good teacher provide the clarity. Thank you. I very much enjoy your channel. 👍👍 Please keep doing this. Next stop for me is the epsilon / delta limit proofs - was always a brain twister for me but I know you will explain it perfectly.

factor out the 5^x: distribute the 1/x power to the 5^x(1 + (4/5)^x) the 5^x/x is just 5^1 = 5 so push that out of the limit. 5 times the limit of (1 + (4/5)^x)^1/x (4/5)^x and 1/x both go to zero as x goes to infinity so that limit simplifies to (1 + 0)^0 = 1^0 = 1 we pushed out the 5 so the final answer is 5 * 1 = 5

You can also think of it another way if you rewrite the base as e^ln(5^x+4^x)^(1/x)=e^(ln(5^x+4^x)/x). By taking the limit as x goes to ∞ in the exponent, it's an indeterminate form of ∞/∞. Take l'Hopital's rule, by taking the derivative of the top and derivative of the bottom. The derivative in the numerator is (5^xln(5)+4^xln(4))/(5^x+4^x). The indeterminate form is still ∞/∞. We can try to keep on going, but the indeterminate form is still ∞/∞, which is leading nowhere up to the point. We have to think outside the box by dividing everything by 5^x both the numerator and denominator. Doing so changes the simplification as (ln(5)+(4/5)^xln(4))/(1+(4/5)^x). We know that (4/5)^x is a decreasing function, so as x goes to ∞, that will go to 0. This means the limit of this exponent is ln(5). Therefore, the limit of that function is e^ln(5)=5.

The "Squeezing theorem" works better here. For x>1 5^x

Limit[(5^x+4^x)^(1/x),x->∞]=5

Great teacher indeed ❤

Thank you,so much❤

Thank you,sir

Great stuff. Thanks. I look forward to checking out your channel. Subscribed. Cheers

What if you took out initially 4^x instead of 5^x. This will give us a limit of 4. You got 5. So we can conclude that the limit does not exist. Just wondering???? 😮

The last ln to bring the power down is not necessary because we do not have indeterminate form we can work directly and he shoukd get full mark not 5/10

Thanks for an other video master

Nice explanation. But log(0) is undefined. So it's wrong to say log(0) = 0. Despite that, lim(y) will still be 1 as supported by the 2nd argument.

3:57 4/5 is less than 1 so its power infinity will be 0. Directly the answer is 5.

Sir I have a method Lim x-infinty (5^x + 4^x)^1/x Now factor out 5^x Lim x-infinity (5^x)^1/x (1 + (4/5)^x )^1/x Now when you put values as 4/5 is less than 1 hence 4/5 power x itend to 0 Hence the final we get 5(1+0)^0 5 answer

At a glance the answer is 5

It could have been solved by intuition, but I liked the rigour.

Was the log really required? Couldn’t we just apply limits after we get 5? The bracket would have become 1.

Finding stationary points of your function using Wolfram Alpha input: solve d/dx{5(1+(4/5)^x)^(1/x)}=0 where x is real returns no real solutions. That means that the function monotonic approaches the limit. Since ((5^1+4^1)^(1/1) = 9 (i.e the value where x=1) and limit =5 then it must be monotonic decreasing to the limit. ↘

Hm, either 1 or 9? Just take the average 🤷🏻‍♂️ works for me

Basic jee mains question

This one was really easy. I'm surprised you covered this but nevertheless, what I did was. Factor out 5^x and you will end up with lim ([5^x(1 + (4/5)^x)]/1^x) Then, distribute the exponents and we end up with -> (5^x)¹/x * (1 + (4/5)^x)¹/x. By one of the many properties of limits that states that the limit of the product is the product of the limits: lim (5^x)¹/x * lim (1 + (4/5)^x))¹/x which you can plug in infinity and you get exactly 5. Never stop learning.

That is amazing! So, the result of "lim (m^x + n^x)^1/x" when m != n, it is max(m, n) , and if m>n, it is m; when m == n, it becomes the double(m or n), it is 2m (or 2n); It feels like (when n tends to m), when n reaches the point of m, the result suddenly goes from m to 2m ~~ ** Can we assert that continuity is only a macro concept? In the microscopic world, they are actually discontinuities? ** ==== Sorry, forgot 2^1/x is also 1.

May I ask why you rewrote the answer as 5? I did get to e^0 = 1 but the last part… I didn’t understand that, please?

u would have said that (5^x+4^x) at x-infi is 5^x so the answer is 5

Wt a chad

but what if we factored out the 4^x?

This just not seem correct though. what you have on the inside is justified as the lim of (1+1/n)^n as x->infinity is the definition of e. You need to rewrite that inner part in terms we can handle. it seems you got a 0*0 situation that is indeterminate and the ln of the limit of y is not justified.

On 2nd thought, I realised that this problem is really easy since the limit of 5^x is infinitely much larger than 4^x as x goes to infinity. For all intent amd purpose, it is zero. So it simplifies as 5^X^(1/x) which is 5 and that’s the answer. I wasn’t in math mode for the first minutes…

I feel like you did many useless work. I think you could have just computed it when you had 5*(1+(4/5)^x)^1/x. 1/x is going to 0. (4/5)^x is also going to 0 as 4/5

Pin plz ❤

At 10:00 the limit of ln 1/x is shown to be zero which is not the case. The correct result is minus infinite. Despite of this the final result is correct.