Basic Limits

Calculus Level 3

f ( x ) = x 5 a + 1 \large f(x) = x^{5a+1}

Function f ( x ) f(x) is defined as above, where a = lim ⁡ n → − 2 2 ( n 3 − 2 n + 4 ) n 3 + 3 n 2 + n − 2 \displaystyle a = \lim_{n \to -2} \frac {2(n^3-2n+4)}{n^3+3n^2+n-2} .

If d 100 f ( x ) d x 100 = y ! \dfrac {d^{100}f(x)}{dx^{100}} = y! at x = 1 x=1 , find y y .


The answer is 101.

Lavisha Parab by Lavisha Parab , 24, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Lavisha Parab
Mar 8, 2015

for a = l i m n → − 2 2 ( n 3 − 2 n + 4 ) n 3 + 3 n 2 + n − 2 a = \underset { n\rightarrow -2 }{ lim } \frac { 2({ n }^{ 3 }-2n+4) }{ { n }^{ 3 }+{ 3n }^{ 2 }+n-2 }

Factorizing we get, a = l i m n → − 2 2 ( n 2 − 2 n + 2 ) ( n + 2 ) ( n 2 + n − 1 ) ( n + 2 ) a=\underset { n\rightarrow -2 }{ lim } \frac { 2({ n }^{ 2 }-2n+2)(n+2) }{ ({ n }^{ 2 }+n-1)(n+2) }

Thus a = 2 ( n 2 − 2 n + 2 ) ( n 2 + n − 1 ) a t n = − 2 a=\frac { 2({ n }^{ 2 }-2n+2) }{ ({ n }^{ 2 }+n-1) } \quad at\quad n=-2

a = 20

d 100 ( x 101 ) d x 100 = 101 ! \frac { d^{ 100 }(x^{ 101 }) }{ dx^{ 100 } } =101!

y = 101 \boxed{101}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

This might sound like nit picky hair splitting from me but I think the derivative should look more like 101 ! x 101!x , and only when it is evaluated at x = 1 x=1 do you obtain 101 ! . 101!. If for example, you evaluated the 10 1 s t 101^{st} derivative at x = 102 , x=102, then the answer would have been y = 102. y=102.

Akeel Howell - 4 years, 7 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...