# Limit to Infinity

I have a question to compute a limit, i understand that the limit is infinity but How can i show the solution to get this answer to infinity. I understand that it will be large pos #'s on top and small #'s on bottom making it go to infinity but how can i write it out to prove this? At least how would i make it look on a test.

$$\lim_{x\rightarrow \ \frac{1}{2}^+} \frac{10+x}{(2x-1)^3}$$

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You can, for example, remark that :$$(1) \quad \forall x \in \left(\frac 12,1 \right) \quad \frac{10+ x}{(2x-1)^3 }\geq \frac{1 }{2x-1}$$ since $(2x-1)^3 \leq 2x-1$ because $0 < 2x-1 < 1$

From $(1)$ , since $$(2) \quad \displaystyle \lim_{x \to \frac{1}{2}^+} \frac1{2x-1}=+\infty$$, we can deduce the rest.

Here is a proof to $(2)$ using the definition of limit (test).

We must proof : $$(3) \quad \quad \forall A>0 \quad \exists \alpha >0 \quad \frac 12 < x < \frac 12 + \alpha \Rightarrow \frac{1}{2x-1} > A$$

It holds since we remark that : $$\frac 1{2x-1} > A \Leftrightarrow 2x-1 < \frac 1A \Leftrightarrow x < \frac 12 + \frac 1{2A}$$ then $\alpha =\frac 1{2A}$ is a convenant response making $(3)$ true.

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Since the denominator will stay positive but go to $0$, and the numerator will be a positive number different from $0$, the quotient will always be positive.

As $x$ gets closer and closer to $\frac{1}{2}$, the denominator will get smaller and smaller and smaller. Dividing a number which will be very close to $10.5$ by a number that is getting smaller, and smaller, and smaller, and smaller (but always stays positive) will result in a number that is getting larger, and larger, and larger, and larger. The quotient will grow without bound and stay positive. So $$\lim_{x\to\frac{1}{2}^+}\frac{10+x}{(2x-1)^3} = \infty.$$

For a basic calculus course, I would accept that as a valid answer; your instructor, though, may have different standards. Exactly how to write it so as to satisfy her will depend on what those standards are, and only she can tell you that.

If you need this to be done using $\epsilon$-$\delta$ type arguments, you need to show that for any $N\gt 0$ you can find $\delta\gt 0$ (which may depend on $N$) with the property that any $x$ such that $0\lt x-\frac{1}{2}\lt \delta$ will satisfy $$\frac{10+x}{(2x-1)^3}\gt N.$$ How to figure it out? Note that for any such $x$, we will have $10+x\gt 10$. So it is enough to make sure that $$\frac{10.5}{(2x-1)^3}$$ is larger than $N$. To make this larger than $N$, we need to make sure that $$\frac{1}{(2x-1)^3} \gt \frac{N}{10}.$$ To make sure this is true, it is enough to make sure that $$(2x-1)^3 \lt \frac{10}{N}$$ which requires $$2x-1 \lt \sqrt[3]{\frac{10}{N}}$$ which requires $$2x \lt 1 + \sqrt[3]{\frac{10}{N}}$$ which requires $$x\lt \frac{1}{2}+\frac{1}{2}\sqrt[3]{\frac{10}{N}}.$$ So, let us set $\delta = \frac{1}{2}\sqrt[3]{\frac{10}{N}}$.

If $$0\lt x-\frac{1}{2}\lt \delta = \frac{1}{2}\sqrt[3]{\frac{10}{N}},$$ then $$0 \lt 2x - 1 \lt \sqrt[3]{\frac{10}{N}}$$ so $$0\lt \frac{1}{\sqrt[3]{\frac{10}{N}}} \lt \frac{1}{2x-1},$$ which is the same as $$0 \lt \frac{\sqrt[3]{N}}{\sqrt[3]{10}}\lt \frac{1}{2x-1};$$ hence $$0\lt \frac{N}{10} \lt \frac{1}{(2x-1)^3}$$ which means that we will have $$0\lt N \lt \frac{10}{(2x-1)^3} \lt \frac{10+x}{(2x-1)^3}.$$ Thus, for any $N\gt 0$ there exists $\delta\gt0$ (for example, $\delta = \frac{1}{2}\sqrt[3]{\frac{10}{N}}$, though any smaller $\delta$ will also work), such that if $0\lt x-\frac{1}{2}\lt \delta$, then $\frac{10+x}{(2x-1)^3} \gt N$. This proves that $$\lim_{x\to\frac{1}{2}^+}\frac{10+x}{(2x-1)^3} = \infty.$$

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That's not how you would write it in a test, I suspect. ;-) –  tomasz Jun 22 '12 at 19:13
@tomasz: Depends on the test. For a basic calculus course in which we don't do much proofs, that's how I would explain it on the board and I would accept it on a test. For an advanced calculus or honor's calculus course in which we do use the definition, then of course you are correct. –  Arturo Magidin Jun 22 '12 at 19:23

You need to show that for any number $N$ you can find an $\varepsilon_N$ such that if $x\in(1/2,\varepsilon_N+1/2)$, then $f(x)>N$. Finding it shouldn't be very hard.

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put $y=x-1/2$ and your limit becomes $lim_{y \to 0+}$ $\frac{(y + 21/2)}{8y^3}$. Thus as $y \to 0^+$, numerator tends to some non-zero finite number(21/2 here), while denominator approaches $0$,hence required limit is infinity.

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As you have guessed, the limit is indeed $\infty$. To show this, we want to show that given $M > 0$, there exists $\delta >0$ such that $\forall x\in(1/2,1/2+ \delta)$, we have $$\dfrac{10+x}{(2x-1)^3} > M$$ Hence, our job is to choose $\delta > 0$.

First note that for $x\in(1/2,1/2+ \delta)$, we have $\dfrac{10+x}{(2x-1)^3} > \dfrac1{(2x-1)^3}$.

Now since $x\in(1/2,1/2+ \delta)$, we have $2x- 1 \in (0, 2\delta)$. Hence, $\dfrac1{(2x-1)^3} > \dfrac1{8\delta^3}$.

Now if we choose $\dfrac1{8\delta^3} = M$, we are done. Hence, given $M>0$, choose $\delta = \dfrac1{2\sqrt[3]{M}}$. Then we get that for $$x \in (1/2,1/2+ \delta) \text{ we have }\dfrac{10+x}{(2x-1)^3} > \dfrac1{(2x-1)^3} > \dfrac1{8\delta^3} > M$$ Hence, given any large positive number $M$, choosing $\delta = \dfrac1{2\sqrt[3]{M}}$ gives us $\dfrac{10+x}{(2x-1)^3} > M$, $\forall x \in (1/2,1/2+\delta)$. Hence, we can conclude that $$\lim_{x \to (1/2)^+}\dfrac{10+x}{(2x-1)^3} = \infty$$

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