#Limits 2

Calculus Level 3

lim x 1 ( 1 + x 2 + x ) 1 x 1 x = a b n \large \lim_{x \to 1} \left(\frac{1+x}{2+x} \right)^{\frac{1-\sqrt x}{1-x}} = \sqrt[n]{\frac{a}{b}}

The equation above holds true for positive integers a a , b b and n n , with a a and b b being primes. Find the value of a + b + n a+b+n .


The answer is 7.

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3 solutions

Chew-Seong Cheong
Mar 29, 2017

L = lim x 1 ( 1 + x 2 + x ) 1 x 1 x = lim x 1 ( 1 + x 2 + x ) 1 x ( 1 x ) ( 1 + x ) = lim x 1 ( 1 + x 2 + x ) 1 1 + x = ( 2 3 ) 1 2 = 2 3 \large \begin{aligned} L & = \lim_{x \to 1} \left(\frac {1+x}{2+x} \right)^{\frac {1-\sqrt x}{\color{#3D99F6}1-x}} \\ & = \lim_{x \to 1} \left(\frac {1+x}{2+x} \right)^{\frac {1-\sqrt x}{\color{#3D99F6}(1-\sqrt x)(1+\sqrt x)}} \\ & = \lim_{x \to 1} \left(\frac {1+x}{2+x} \right)^{\frac 1{1+\sqrt x}} \\ & = \left(\frac 23 \right)^\frac 12 \\ & = \sqrt{\frac 23} \end{aligned}

a + b + n = 2 + 3 + 2 = 7 \implies a + b + n = 2+3+2 = \boxed{7}

Rahil Sehgal
Mar 29, 2017

why you do to complicated way if it can done simply?? But nice one

Nivedit Jain - 4 years, 2 months ago

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I was trying a new way to solve such a problem..

Rahil Sehgal - 4 years, 2 months ago

Good!, Did it like this. I ddint knew FIITJEE is so advanced...

You know all concepts of XIth and XIIth ?

Md Zuhair - 4 years, 2 months ago

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Thanks brother...:) Keep uploading questions in future

Rahil Sehgal - 4 years, 2 months ago

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You know all concepts of XIth and XIIth ?

Md Zuhair - 4 years, 2 months ago

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@Md Zuhair Not all... But some...

Rahil Sehgal - 4 years, 2 months ago
Andrea Virgillito
Mar 30, 2017

I solved It mentally: the only problem is at the exponent but it is in the form (1-x)/(1-x^2) and by semplify and substituting we get to (2/3)^ 1/2 2+2+3=7

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