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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?

Xander Henderson
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VictorVH
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  • Is $f$ defined at $0$? Is it continuous at $0$? Can you express it as a quotient of two continuous functions $f(x) = g(x) / h(x)$ where $h(0) \ne 0$? – David Kraemer Feb 11 '19 at 17:21

4 Answers4

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Start with $| \frac{2x+3}{x+1}-3|=|\frac{-x}{x+1}|$.

For $|x|< \delta$, you have $|-x|=|x|<\delta$.

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

Put everything together to get: $|\frac{-x}{x+1}|<\frac{\delta}{1-\delta}$, and choose $\delta$ to be less than $\frac{\epsilon}{1+\epsilon}$.

Locally unskillful
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  • Why is it that we choose $\frac{\epsilon}{1+\epsilon}$ when the $\delta$ says $\frac{\delta}{1-\delta}$ ? Might be a stupid question, but I don't understand the conclusion. – VictorVH Feb 11 '19 at 17:50
  • You wish to have $\frac{\delta}{1-\delta}< \epsilon$. Now rearrange to get $\delta< \frac{ε}{1+ε}$ – Locally unskillful Feb 11 '19 at 17:54
  • Oh thanks, I guess my understanding of this is not quite there yet. More practice I guess. – VictorVH Feb 11 '19 at 17:57
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You need to show that $\left \lvert \frac{2x+3}{x+1} - 3\right \rvert$ can be made arbitrarily small, by taking $x$ close to $0$.

Notice that $$\left \lvert \frac{2x+3}{x+1} - 3\right \rvert = \frac{\lvert x\rvert}{\lvert x + 1\rvert}.$$ Now when $\lvert x \rvert < 1/2$, we have $\lvert x +1 \rvert > 1/2$ and so $\frac{1}{\lvert x +1\rvert} < 2$. Thus for $\lvert x \rvert < 1/2$, we see $$\left \lvert \frac{2x+3}{x+1} - 3\right \rvert = \frac{\lvert x\rvert}{\lvert x + 1\rvert} < 2\lvert x \rvert.$$ Fix $\epsilon > 0$. We need to choose $\delta$ so that $\lvert x \rvert < \delta$ ensures that $\left \lvert \frac{2x+3}{x+1} - 3\right \rvert < \epsilon.$ The above equation suggests we take $\delta = \epsilon/2$, since then $\lvert x \rvert < \delta$ implies that $$\left \lvert \frac{2x+3}{x+1} - 3\right \rvert = \frac{\lvert x\rvert}{\lvert x + 1\rvert} < 2\lvert x \rvert < \epsilon.$$ However, the first inequality only holds when $\lvert x \rvert < 1/2$, so we also need to ensure that this holds. Hence we take $\delta = \min\{1/2, \epsilon/2\}$, and then we get the desired conclusion.

User8128
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  • Hey I understand everything in this proof except the end where you take the minimum. Aren't $\delta$ supposed to be dependent on $\epsilon$? How can we then set $\delta = 1/2$? – VictorVH Feb 11 '19 at 17:53
  • You can’t just set $\delta =1/2$, you have to set $\delta = \min{1/2,\epsilon/2}$. This still very much depends on $\epsilon$. If $\epsilon =0.1$ then you have to set $\delta= 0.05$. However if $\epsilon = 4$, then (to follow this proof), you have to set $\delta = 1/2$. – User8128 Feb 11 '19 at 18:58
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$|\dfrac{2x+3}{x+1}-3|=|\dfrac{-x}{x+1}|=\dfrac{|x|}{|x+1|}$.

For $|x|<1/2$, we have

$-1/2<x<1/2$, and

$1/2 <x+1<3/2$,

Let $\epsilon >0$ be given.

Choose $\delta =\min(1/2,\epsilon/2)$.

Then $|x| < \delta$ implies

$|\dfrac{2x+3}{x+1}-3|=\dfrac{|x|}{|x+1|}<\dfrac{\delta}{1/2}=$

$2\delta \le \epsilon.$

Peter Szilas
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if $\lim_\limits {x\to a} f(x) = l$ and $\lim_\limits{x\to a} g(x) = m$ and $m\ne 0$ then $\lim_\limits{x\to a} \frac {f(x)}{g(x)} = \frac {l}{m}$

$\forall \epsilon_1 > 0, \exists \delta >0 : |x-a|<\delta_1 \implies |g(x) - {m}|<\epsilon_1$

$x<\delta_1 \implies |g(x)| > 0$

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

$\forall \epsilon_2 > 0, \epsilon_1 < \frac 1M \implies |\frac {1}{g(x)} - \frac 1{m}|<\epsilon_2$

$\forall \epsilon_3 > 0, \exists \delta_2 >0 : |x-a|<\delta \implies |f(x) - l|<\epsilon_3$

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

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

Doug M
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