$$\lim_{x\rightarrow 0} \frac{\frac{1}{x+3}-\frac{1}{3}}{x}$$
Is the limit $-\displaystyle{\frac{1}{9}}$?
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First combine the numerator, it doesn't even need l'Hospital:
$\lim_{x \rightarrow 0} \frac{\frac{3-x-3}{3x+9}}{x}=\lim_{x \rightarrow 0} \frac{\frac{-x}{3x+9}}{x}=\lim_{x \rightarrow 0} \frac{\frac{-1}{3x+9}}{1}$
And here is already clearly visible the result.
peterh
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$$\lim_{x\rightarrow 0} \frac{\frac{1}{x+3}-\frac{1}{3}}{x}$$ Lets denote $f(t)=\frac{1}{t}$ $$\lim_{x\rightarrow 0} \frac{f(3+x)-f(3)}{x}$$ Now wait a second, it looks familiar $$f'(3)\equiv \lim_{x\rightarrow 0} \frac{f(3+x)-f(3)}{x}$$ And we know that $$f'(t)=\frac{-1}{t^2}$$ Thus $$\lim_{x\rightarrow 0} \frac{\frac{1}{x+3}-\frac{1}{3}}{x}=\frac{-1}{9}$$ This method should not be considered a "valid solution", since the derivative is defined by the limit, but it's a nice way look at it
Uri Goren
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Use \lim in math mode to produce $\lim$. Also, $$f'(3) = \lim_{h \to 0} \frac{f(3 + h) - f(\color{red}{3})}{h}$$ Here, $h$ is $x$. – N. F. Taussig Sep 22 '15 at 10:06
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$$\frac{\dfrac1{x+3}-\dfrac13}x=-\frac1{3(x+3)}.$$ You should be able to conclude.
Lim [(1-x/3)-(1/3)]/x x-->0, which is $\displaystyle\lim_{x\to0}\dfrac{\left(1-\frac{x}{3}\right)-\frac{1}{3}}{x}$, which does not exist (as a two-sided limit). Please clarify. – dbanet Sep 20 '15 at 16:511/x+3is taken to mean $\frac{1}{x}+3$, not $\frac{1}{x+3}$. If you want $\frac{1}{x+3}$ and you do not properly typeset using TeX or MathJax, then you must use parentheses for the denominator as1/(x+3). – JMoravitz Sep 20 '15 at 16:55