Limit problem

[Mehgan]

Hi, can anyone show me how to answer this problem?

lim
x->0 (4+x)^-1 -4^-1
x

I would be much obliged!

 

[Mehgan]

oops, that bottom x was meant to be the denominator for the fraction. Sorry y'all.

 

[Unco]

\(\displaystyle \mbox{ \lim_{x\to 0} \frac{(4 + x)^{-1} - 4^{-1}}{x}}\)

Rewrite the numerator as 1/(4+x) - 1/4, and subtract the fractions (by cross-multiplying, etc); then you can simplify.

 

[nbacardi]

answer: lim ((1/(4+x))-1/4).1/x

= (1/(4x+x^2))-(1/4x) = (1-(1+x/4))/(4x+x^2) = (x/4)/(4x+x^2)

= x/(4(4x+x^2)) = x/(16x+4x^2) = x/(x(16+4x))

then the deno and num x gets canceld, so it becomes
= 1/(16+4x) and then just replace the x by 0 and you answer is = 1/16

good luck!
Take care

 

[Mehgan]

ok, thanks!

 

[Unco]

Nbacardi lost a negative sign, by the way.