Jee Mains pyq on L'Hôpital's Rule #jeemains #limit #pyq #maths

Struggling with limit problems in JEE Mains? 😰 Don't worry, I've got you covered! In this video, we dive into the magic of L'Hôpital's Rule with Previous Year Questions (PYQs) to help you ace your exams. 🚀 Learn tips, tricks, and the best approach to tackle tricky limit questions effortlessly.

📚 Topics Covered:

When and how to use L'Hôpital's Rule ✅
Common mistakes to avoid ⚠️
Step-by-step solutions of PYQs 🎯
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#JEEMains2024 #Limits #LHôpitalRule #MathsPYQ #IITJEE

[bourbaki-]

Since, the denominator tends to 0, for existence of the limit we need the numerator must tends to 0.
Hence, we have 3+asinx+bcosx+ln(1-x) = 0 at x=0.
So, 3+b = 0 => b = -3
And by using L'Hopital similarly by existence of limit, we need acosx-3sinx-1/(1-x) = 0 at x = 0
So, a-1 = 0 => a = 1.
Hence, 2a-b = 5

[castor]

Limit is just the tendiveness even small fnal changes can flip da game ❤❤

[TheTECHVEDA]

Sir!! the series like this are very helpful for us, please bring more videos as above. We are really loving it

[swetapadmamahapatra]

Answer is 5. a = 1 and b = -3
Let f(x)=3+asinx +bcosx +ln(1-x)
And let g(x) be 3tan²x.
Given Limit of f(x)/g(x) as x tends to 0=1/3.
But g(x)= 3tan²x becomes zero.
If f(x) is not equal to zero,
i.e, f(x)<0 or greater than 0,
Then, the limit will approach, -∞ or ∞ respectively.
So for the limit to be equal to 1/3.
So f(x)/g(x) must be in an indeterminate form.
Here in this case the indeterminate form would be 0/0.
So 3+asinx +bcosx +ln(1-x)=0
As x approaches zero.
=> 3+b +ln (1)=0
=> 3+b +0=0
=> b= -3.
Now, by applying L' Hospitals rule,
We get Lim f'(x)/g'(x)= 1/3
(As x approaches zero)
=> Lim of (acosx +3sinx -(1/1-x))/(6tanx.sec²x) = 1/3.

Again as x approaches 0, g'(x)=0
So for the limit to be equal to 1/3.
f'(x)/g'(x) must be in 0/0 form

=> f'(x)=0 (as x approaches zero)
=> acosx -bsinx -1/1-x =0
=> a cos (0) - bsin (0) -1/1-0=0
=> a -1 = 0
=>a= 1
=> 2a -b = 2×1 -(-3)= 2+3=5
Or by multiplying 3 on both R. h.s and L.H.S,
One gets
Lim 3+asinx +bcosx + ln(1-x) /tan²x =1
Bu differentiating it the second time,
We get -asinx -bcosx -1=2
=> -b=3
=> b=3.

Thank you sir 💜

[devtomer]

Thank you bhaiya, , mere se ye ho nahi rha tha aur online koi L hospital se ker hei nahi rha tha, ,, aap ne ker diya shukriya ❤

[Letscrrkkit]

Please start solving good pyqs in shorts like 'divine jee'

[adityajha]

Binomial bhi use krskte the expansion likh k

[Ashutosh-mdwt]

5 is answer 1 wala achha question tha❤❤❤

[KrishnaKantaBehera-rggk]

Sir yahan pe toh phasa diya aapne 😢
Mujhe thodi yaad thaa Taylor series jaisa bhi kuch thaa
Woh toh itna notebook ragadne ke baad kahin pe likha hua thaa ki 💀
sinx, cosx aur tanx ke liye special expansions thee bade bade wale

Phir bhi option(C) 5 aayega shayad

question mein diya hai
limˣ ⃗ ⁰ 2(jo bhi likha hua hai)/3tan²x = 1/3

Abhi ke liye teeno expansions kaa use karna sehat ke liye haanikarak hai 😂
Toh ek kaam karenge
tan²x ke saath x² multiply aur divide kardenge
toh limˣ ⃗ ⁰ tan²x/x² toh one hoo jaayega
toh niiche mein sirf 3x² rahega

limˣ ⃗ ⁰ 2(jo bhi likha hua hai)/3x² = 1/3
ab Taylor Series ke hisab se
cosx = 1-x²/2 +x⁴/24 -
sinx = x - x³/6 + x⁵/120
(Main shayad log mention karna bhul gaya, uske liye bhi alag expansion hai 😅)
log(1-x) = -x - x²/2 - x³/3

Ab hum ye saara raita daal denge lim mein 💀 (type karte karte chal basunga)
lim ˣ ⃗ ⁰ [3 +α(x - x³/6 + x⁵/120 + β(1-x²/2 +x⁴/24 - + (-x - x²/2 - x³/3

Ek example dena chahunga (pata nahi thik hai ki nahi)
lim ˣ ⃗ ⁰ 1/x undefined hota hai

toh ye pura function tab hi defined hoga jab numerator mein sirf x² aur usse upar wale terms rahenge
toh jo "x" ke coefficients aur constant terms honge toh unko eliminate hona padega
(Agar nahi honge toh 1/x aur 1/x² aayega jisme limit put karne pe undefined hoga)

lim ˣ ⃗ ⁰ [3 +α(x - x³/6 + x⁵/120 + β(1-x²/2 +x⁴/24 - + (-x - x²/2 - x³/3
Isme se hum "x" ke coefficients aur constant terms alag karlenge

lim ˣ ⃗ ⁰ [(3 + β) + x(α - 1) + α(- x³/6 + x⁵/120 + β(-x²/2 +x⁴/24 - + ( - x²/2 - x³/3

(Ab ye mat puuchna ki alag kaise kiya... pehle se myopia hai ab hypermetropia bhi hoo jaayega)
(3 pehle se hai aur cosx ke expansion mein pehla term 1 thaa toh β se multiply hoo gaya)
(log(1-x) ke expansion mein pehle term aur sinx ke expansion mein pehle term dono mein hi "x" the toh alag kar diye)

Ab jaise ki mention kiya
toh jo "x" ke coefficients aur constant terms honge toh unko eliminate hona padega taaki function defined hoosake
(warna toh hehe)

toh ab
3+β = 0, α -1 = 0
ab isme daal denge
2α - β
2 +3
5 is the answer

[ArnavSingh-eglu]

5 is the answer i solved this question today itself

[AAAaAAae]

can use expansions of sinx, cosx and log x

[ShubhamYadav-efpn]

Cos ka term 1 kaise?
(1-cosx)/x wali identity se n?

[codblaze]

Sir main chapters bta dijiye jiske 100% question aayenge

[ashishrajsingh]

Sir appne log kyu nahi liya?? Easily ho jata

[subhodippaul]

I know a better way to solve this question