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Phrases Limit PYQ








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limx1x41x1=limxkx3k2x2k2=, then find k





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Solution

Quick DL Method Solution

Given:

limx1x41x1=limxkx3k2x2k2

LHS using derivative:

limx1x41x1=ddx(x4)|x=1=4x3|x=1=4

RHS using DL logic:

limxkx3k2x2k23k2(xk)2k(xk)=3k2

Equating both sides:

3k2=4k=83

k=83



Let f(x)=x21|x|1. Then the value of limx1f(x) is





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Solution



 is equal to 





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Solution



The value of the limit limx0(1x+2x+3x+4x4)1/x
is





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Solution



Which of the following is NOT true?





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Solution



For aR (the set of al real numbers), a1, limn(1a+2a++na)(n+1)a1[(na+1)(na+b)(na+n)]=160 . Then one of the value of a is





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Solution



The value of Ltx0exex2x1cosx is equal to 





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Solution

Evaluate: limx0exex2x1cosx

Step 1: Apply L'Hôpital's Rule (since it's 0/0):

First derivative: ex+ex2sinx

Still 0/0 → Apply L'Hôpital's Rule again: exexcosx

Now, limx0111=0

Final Answer: 0



limx(x+7x+2)x+5 equal to





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Solution



Let f(x) be a polynomial of degree four, having extreme value at x = 1 and x = 2. If limx0[1+f(x)x2]=3, then f(2) is





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Solution

Given it has extremum values at x=1 and x=2
⇒f′(1)=0  and  f′(2)=0
Given f(x) is a fourth degree polynomial 
Let  f(x)=ax4+bx3+cx2+dx+e
Given 
limx0[1+f(x)x2]=3
limx0[1+ax4+bx3+cx2+dx+ex2]=3
limx0[1+ax2+bx+c+dx+ex2]=3
For limit to have finite value, value of 'd' and 'e' must be 0
⇒d=0  & e=0
Substituting x=0 in limit 
⇒ c+1=3
⇒ c=2
f(x)=4ax3+3bx2+2cx+d
x=1 and x=2 are extreme values,
f(1)=0 and $f^{\prime}(2)=0
4a+3b+4=0 and 32a+12b+8=0 
By solving these equations
we get, a=12 and b=2
So,
f(x)=x422x3+2x2
f(x)=x2(x222x+2)
f(2)=22(24+2)
f(2)=0



The value of limxaa+2x3x3a+x2x





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Solution



What is the value of limx(x+1)(e1x+11)?





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Solution








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Solution

Function is the form of  therefore using by L'Hospital rule
Again apply L'Hospital Rule,
Putting x = 0, we get 
 



Find 





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Solution



If f(x)=limx06x3x2x+1loge9(1cosx) is a real number then limx0f(x)





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Solution



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