$\epsilon - \delta$ proof. Need help I have been knocking my head against this proof for an hour now and I'm no closer to solving it. I simply have no idea where to go. The thing I need to prove is, given
$$f(x)=\frac{2x+3}{x+1},$$
show that $f(x)\to 3$ for $x\to 0$.

I have tried to factor it out but I simply have no luck. Can anybody give me a hint of where to go with this?
Start with $| \frac{2x+3}{x+1}-3|=|\frac{-x}{x+1}|$.

For $|x|&lt; \delta$, you have $|-x|=|x|&lt;\delta$.

For the denominator, use the reverse triangle inequality to get:
$|1+x| \geq 1 - |x| &gt;1-\delta$.

Put everything together to get: $|\frac{-x}{x+1}|&lt;\frac{\delta}{1-\delta}$, and choose $\delta$ to be less than $\frac{\epsilon}{1+\epsilon}$.
if $\lim_\limits {x\to a} f(x) = l$ and $\lim_\limits{x\to a} g(x) = m$ and $m\ne 0$ then $\lim_\limits{x\to a} \frac {f(x)}{g(x)} = \frac {l}{m}$

$\forall \epsilon_1 &gt; 0, \exists \delta &gt;0 : |x-a|&lt;\delta_1  \implies |g(x) - {m}|&lt;\epsilon_1$

$x&lt;\delta_1 \implies |g(x)| &gt; 0$

let $M= \inf \{|g(x)|: |x-a|&lt;\delta_1|\}$

$\forall \epsilon_2 &gt; 0, \epsilon_1 &lt; \frac 1M \implies |\frac {1}{g(x)} - \frac 1{m}|&lt;\epsilon_2$

$\forall \epsilon_3 &gt; 0, \exists \delta_2 &gt;0 : |x-a|&lt;\delta  \implies |f(x) - l|&lt;\epsilon_3$

$\forall \epsilon &gt; 0, \exists \delta &gt;0 : |x-a|&lt;\delta_1  \implies |\frac {f(x)}{g(x)} - \frac {l}{m}|&lt;\epsilon$

$|f(x) \frac {1}{g(x)} - l\frac {1}{m}|\\
|f(x) \frac {1}{g(x)} -f(x)\frac {1}{m} + f(x)\frac {1}{m}- l\frac {1}{m}|\\
|f(x) (\frac {1}{g(x)} -\frac {1}{m}) + (f(x)-l)\frac {1}{m}| &lt; |f(x) |(\frac {1}{g(x)} -\frac {1}{m})| + |(f(x)-l)|\frac {1}{m}| &lt; |f(x)|\epsilon_2 + |\frac {1}{m}|\epsilon_3$