$\large \displaystyle \lim_{x \rightarrow 0} \left ( \sum_{n = 0}^{\infty} x^n \right )$

Find the value of the expression above.

0
Indeterminate
None of the above
1

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Simple standard approach.

**
Bonus question
**
: Can you solve this by Big O Notation?

You know you just derived the sum for infinite geometric progressions formula?

Sharky Kesa
- 6 years ago

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Yeah! You are right.

Swapnil Das
- 6 years ago

The "derivation" here assumes convergence rather than showing the radius (interval) of convergence which is $0\lt |x|\lt 1$ . So, I wouldn't call it a complete proof / derivation.

Prasun Biswas
- 6 years ago

**
[Response to Challenge Master Note]
**

By "solve this using Big O Notation", did you mean the following?

$\begin{aligned}\lim_{x\to 0}\sum_{n=0}^\infty x^n&=\lim_{x\to 0}\bigg(x^0+\mathcal{O}(x)\bigg)\\&=\lim_{x\to 0}\bigg(1+\mathcal{O}(x)\bigg)\\&=\left(\lim_{x\to 0}1\right)+\left(\lim_{x\to 0}\mathcal{O}(x)\right)=1+0=1\end{aligned}$

Here, $\mathcal{O}(\cdot)$ is the Big O Notation.

Prasun Biswas
- 5 years, 12 months ago

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How do you know that $\displaystyle \lim_{x\to 0} x^0 = 1$ ?

Pi Han Goh
- 5 years, 12 months ago

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Initially the limit is not put anything to the power 0 is considered as 1 . And then the limit is put to consideration which is uneffective with the absence of x

Shivam Kalra
- 5 years, 10 months ago

Graph of x^0 At 0- lhl=1 At 0+ rhl=1

Ameya Bhamare
- 5 years, 10 months ago

If I recall correctly, $x^0=1~\forall~x\in\Bbb{C}\setminus\{0\}$ by definition. Isn't that a sufficient reason to conclude that the two-sided limit at $x=0$ exists and is equal to $1$ ?

Prasun Biswas
- 5 years, 12 months ago

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@Prasun Biswas – If you're using definitions, then there's nothing else to prove. I thought we need to use some epsilon delta proof. No?

Pi Han Goh
- 5 years, 12 months ago

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@Pi Han Goh – Maybe we can take $\delta=\epsilon$ for the epsilon delta proof? I'm not sure though. I haven't yet studied the epsilon-delta proof technique properly.

Although, I can't see how we can give even an epsilon-delta proof without using the definition stated above because I can't see any way of getting a suitable $\delta$ otherwise.

I'm stuck at $|x^0-1|\lt\epsilon$ (while working backwards to find a suitable $\delta$ ).

Do enlighten me if you have some sort of proof that avoids using that definition.

Prasun Biswas
- 5 years, 12 months ago

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@Prasun Biswas – What I did was set $y = \frac1x$ , then the limit becomes $\displaystyle\lim_{y\to\pm\infty} \frac1{y^0} = \frac1{1} = 1$ because any number (other than 0) raise to the power of 0 yields 1.

Pi Han Goh
- 5 years, 12 months ago

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@Pi Han Goh – That isn't much different from what I did. Your method also used the definition of $x^0$ .

In my method, we have $x\to 0\implies x\neq 0$ .

Hence, our approaches are ultimately the same.

Prasun Biswas
- 5 years, 12 months ago

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@Prasun Biswas – Yours further implies that the two-sided limit at $x=0$ exist and equals to $1$ , while mine didn't. I didn't assume that the limit must exist at $x=0$ .

Pi Han Goh
- 5 years, 12 months ago

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@Pi Han Goh – Ah yes, I agree. You showed that both the left and right sided limit exists and equals $1$ which implies that the two-sided limit at $x=0$ exists and equals $1$ . Yes, doing that is better. But I thought that it was fairly obvious that the two-sided limit exists at $x=0$ when we use that definition of $x^0$ .

Prasun Biswas
- 5 years, 12 months ago

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@Prasun Biswas – Your turn! Prove that $\displaystyle\lim_{x\to0} 0^x$ does not exist.

Pi Han Goh
- 5 years, 12 months ago

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@Pi Han Goh – Again, it's obvious that $0^x=0~\forall~x\in\Bbb{R^+}$ . Hence, we have,

$\lim_{x\to 0^+} 0^x=\lim_{x\to 0^+} 0=0\\ \lim_{x\to 0^-} 0^x=\lim_{x\to 0^+}\frac 1{0^x}=\textrm{(undefined)}$

Since the left-hand limit at $x=0$ doesn't exist (is undefined), the given two-sided limit doesn't exist.

I have no good problems to give you since they'll all be easy for you. :|

Prasun Biswas
- 5 years, 12 months ago

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@Prasun Biswas – In the original question the first term of the series is $\displaystyle \lim_{x\to 0} x^0$ because n=0.

$\large x^{0} = e^{ln(x^{0})} = e^{0 \cdot ln(x)} = e^{0} = 1$

So the limit of the series is 1.

Brandon Stocks
- 5 years, 1 month ago

Amazing, No? The Things we do unknowingly :P

Mehul Arora
- 6 years ago

Good solution

Fahad Shah
- 5 years, 5 months ago

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$\displaystyle \left ( \sum_{n = 0}^{\infty} x^n \right )$

$=1+x+x^2+x^3+x^4.........$

Let $y=1+x+x^2+x^3+x^4........$

$\rightarrow y-1=x+x^2+x^3........ \\ \rightarrow y-1=x(1+x+x^2+x^3....) \\ \rightarrow y-1=xy \\ \rightarrow y-xy=1 \\ \rightarrow y(1-x)=1 \\ \rightarrow y=\frac{1}{1-x} \\ \rightarrow\large \displaystyle \lim_{x \rightarrow 0} \left ( \sum_{n = 0}^{\infty} x^n \right ) = \displaystyle \lim_{x \rightarrow 0} (\frac{1}{1-x}) = \frac {1}{1-0} = \boxed{1}$

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