Question

I need help find this limit analytically . Each step would help alot .

[1/(3+x)]-(1/3)
lim ------------
X-->0 X

Answer 1

The formula can be rearranged to
\[-\frac{x}{3x^2+9x}\]
Now when x equal 0, so do both numerator and denominator, which means we are able to apply L'Hopital's rule which states that the limit of the original fraction is equal to the limit of the derivative as x->0.
Algebraically:
\[\lim_{x \rightarrow 0}(-\frac{x}{3x^2+9x}) = lim_{x \rightarrow 0}(-\frac{1}{6x+9})\]
Which evidently tends to \[-\frac{1}{9}\]

Answer 2

ooh did you foil and elimate to get it all simplified?

Answer 3

i havent been taught deritives either.Do i need to know it for this problem?

Answer 4

Here's the simplification:
\[\frac{\frac{1}{x+3} - \frac{1}{3}}{x}\]
\[= \frac{1}{(x+3)x} - \frac{1}{3x}\]
\[= \frac{1}{x^2+3x} - \frac{1}{3x}\]
\[= \frac{3x - x^2+3x}{3x^3+9x^2}\]
\[= \frac{-x^2}{3x^3+9x^2}\]
\[= \frac{-x}{3x^2+9x}\]
I'm surprised you've been given a question like this if you haven't been taught derivatives.

Answer 5

Sorry slight mistake with one of the signs in the fourth line.
It should be:
\[\frac{3x - x^2-3x}{3x^3+9x^2}\]

Answer 6

thanks alot i think i can figure it out now.