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Calculus Level 2

Solve lim ⁡ k → 100 k 100 + 100 k − ( 100 100 + 100 × 100 ) k − 100 \lim _{ k\rightarrow 100 }{ \frac { { k }^{ 100 } + 100k - ({ 100 }^{ 100 }+100\times 100) }{ k - 100 } }

∞ \infty 0 0 100 100 { 100 }^{ 100 } − ∞ -\infty 100 99 { 100 }^{ 99 } 100 99 + 100 { 100 }^{ 99 } + 100 Indeterminate 100 ( 10 99 + 10 0 ) 100 ({ 10 }^{ 99 } + { 10 }^{ 0 })

Abyoso Hapsoro by Abyoso Hapsoro , 22, Indonesia

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1 solution

Tijmen Veltman
May 29, 2015

Substitute k = x + 100 k=x+100 :

lim ⁡ x → 0 ( x + 100 ) 100 + 100 ( x + 100 ) − ( 10 0 100 + 100 × 100 ) x \lim_{x\to 0} \frac{(x+100)^{100}+100(x+100)-(100^{100}+100\times 100)}{x}

= lim ⁡ x → 0 O ( x 2 ) + 100 × 10 0 99 x + 10 0 100 + 100 x + 10 0 2 − 10 0 100 − 10 0 2 x =\lim_{x\to 0} \frac{\mathcal{O}(x^2)+100\times 100^{99}x+100^{100}+100x+100^2-100^{100}-100^2}{x}

= lim ⁡ x → 0 100 × 10 0 99 + 100 + O ( x ) =\lim_{x\to 0} 100\times 100^{99}+100+\mathcal{O}(x)

= 100 ( 1 0 99 + 1 0 0 ) =\boxed{100(10^{99}+10^0)} .

The big O notation O ( x k ) \mathcal{O}(x^k) in this case stands for any polynomial terms of order k k or higher.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

it is 100^99

not 10^99

all options are wrong

Anand O R - 5 years, 8 months ago

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