Limits 0/0

Calculus Level 4

lim ⁡ x → 1 f ( x ) x − 1 = 2 , lim ⁡ x → 1 [ g ( x ) ( 5 x + 4 − 4 x 2 + 5 ) ] = − 1 \lim _{ x\to1 }{ \frac { f( x ) }{ \sqrt { x } -1 } } =2 , \qquad \lim _{ x\to 1 }{ \left[ g( x ) (\sqrt { 5x+4 } -\sqrt { 4{ x }^{ 2 }+5 } ) \right] }=-1

Let f ( x ) f(x) and g ( x ) g(x) be functions satisfying the equations above. Find lim ⁡ x → 1 [ f ( x ) g ( x ) ] \displaystyle \lim_{x\to1} [ f(x) g(x) ] .


The answer is 2.

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Μπαμπινος Μ.Σ by Μπαμπινος Μ.Σ , 21, Greece

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2 solutions

1 Helpful 0 Interesting 1 Brilliant 0 Confused
Sabhrant Sachan
Jul 9, 2016

I totally agree with M π α μ π i v o ζ , But there is a easier way out . Consider lim ⁡ x → 1 f ( x ) x − 1 ⋅ − 1 − 1 = 2 lim ⁡ x → 1 f ( x ) ⋅ g ( x ) ⋅ ( 5 x + 4 − 4 x 2 + 5 ) 1 − x = 2 lim ⁡ x → 1 f ( x ) g ( x ) ⋅ lim ⁡ x → 1 ( 5 x + 4 − 4 x 2 + 5 ) 1 − x = 2 lim ⁡ x → 1 f ( x ) g ( x ) ⋅ lim ⁡ x → 1 1 2 5 x + 4 ⋅ 5 − 1 2 4 x 2 + 5 ⋅ ( 8 x ) − 1 2 x = 2 lim ⁡ x → 1 f ( x ) g ( x ) = 2 \text{I totally agree with M} \pi \alpha \mu \pi i v o \zeta , \text{But there is a easier way out . Consider} \\ \displaystyle \lim_{x \to 1} \dfrac{f(x)}{\sqrt{x}-1}\cdot \dfrac{-1}{-1} = 2 \\ \displaystyle \lim_{x \to 1} f(x)\cdot g(x) \cdot \dfrac{\left( \sqrt{5x+4}-\sqrt{4x^2+5}\right)}{1-\sqrt{x}} = 2 \\ \displaystyle \lim_{x \to 1} f(x) g(x) \cdot \displaystyle \lim_{x \to 1} \dfrac{\left( \sqrt{5x+4}-\sqrt{4x^2+5}\right)}{1-\sqrt{x}} = 2 \\ \displaystyle \lim_{x \to 1} f(x) g(x) \cdot \displaystyle \lim_{x \to 1} \dfrac{ \dfrac{1}{2\sqrt{5x+4}}\cdot5-\dfrac{1}{2\sqrt{4x^2+5}}\cdot(8x)}{-\dfrac{1}{2\sqrt{x}}} = 2 \\ \boxed{\displaystyle \lim_{x \to 1} f(x) g(x) = 2 }

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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