I have recently come up with a proof of 1/0 = ∞. Here is the proof: We will first write 1 and 0 as powers of 10. 1 = 10^0 Since the limit of 10^x as x approaches -∞ is equal to 0, 0 = 10^-∞ Therefore, 1/0 = (10^0)/(10^-∞) = 10^(0-(-∞)) = 10^∞ = ∞  Is this proof correct?
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My brother was teaching me the basics of mathematics and we had some confusion about the positive and negative behavior of Zero. After reading a few post on this we came to know that it depends on the context of its use.  Why do we take 1/0 as positive infinity rather than negative infinity (we come close to zero from negative axis)?
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Is the fraction  $$\frac{1}{\frac{1}{0}}$$  undefined?   I know that division by zero is usually prohibited, but since dividing a number by a fraction yields the same result as multiplying the number by the fraction's reciprocal, you could argue that  $$\frac{1}{\frac{1}{0}} = (1)\left(\frac{0}{1}\right) = 0$$  Is that manipulation permissible in this case? Why or why not?
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I was trying to find a proof of $a\times0 = 0$ by myself (assuming commutativity, associativity, distributivity, etc) and I came up with $$ a+0=a(1) \implies 1 = \frac{a+0}{a} = \frac aa + \frac 0a = 1 + \frac 0a \implies \frac 0a = 0 \implies a(0)=0$$ Is there anything wrong with this proof, or does it assume anything that shouldn't be assumed?
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$x^x = 100$.   I have no clue on how to solve this. If you guys have, please show me your solution as well.
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Why is ${x^{\frac{1}{2}}}$ the same as $\sqrt x $?   I'm currently studying indices/exponents, and this is a law that I was told to accept without much proof or explanation, could someone explain the reasoning behind this.  Thank you.
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I accept that two numbers can have the same supremum depending on how you generate a decimal representation. So $2.4999\ldots = 2.5$ etc. Can anyone point me to resources that would explain what the below argument that shows $999\ldots = -1$ is about? Here is the most usual proof I see that $0.999\ldots = 1$:  $x=0.999\ldots$ $10x=9.999\ldots$ $10x - x = 9$ $x=1$  Using this same argument template I can show $999\ldots=-1$:  $x= \ldots9999.0 $ $0.1x= \ldots9999.9$ $0.1x - x = 0.9$ $x=-1$  What might this mean? Edit from one of the comments: $$\sum_{k=0}^{\infty}{9 \cdot 10^k}=-1$$
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Can you please explain why \begin{align*} \lim_{x\to 0^+}{x^{x^x}} &amp;= 0 \end{align*}
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In other words, how to prove:  For all real constants $a$ and $b$ such that $a &gt; 1$, $$\lim_{n\to\infty}\frac{n^b}{a^n} = 0$$  I know the definition of limit but I feel that it's not enough to prove this theorem.
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On Wikipedia, it says that $\lim_{n\to\infty}(1+\frac{1}{n})^n = e$ :     It [e] is approximately equal to 2.71828,[1] and is the limit of (1 + 1/n)n as n approaches infinity, ... ()   When I evaluate $(1+\frac{1}{n})^n$ for $n = 10^8$, I get approximately $2.718281798347$ which indeed is pretty close to $e$.  But when I try to "solve" the limit using the laws of limits, I get  $$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = \left(\lim_{n\to\infty}\left(1+\frac{1}{n}\right)\right)^n$$  because of the power law  $$ = \left(\lim_{n\to\infty}(1)+\lim_{n\to\infty}\left(\frac{1}{n}\right)\right)^n$$  because of the addition law  $$=\left(1+0\right)^n = 1$$  but that would mean that $e=1$, which is obviously not true.  What am I missing / doing wrong?  Thanks in advance.
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If a>0 and b>0,  $ \lim_{x\rightarrow +\infty } \sqrt[x]{a^x+b^x} = ? $  What I was trying to do: Suppose a>b. Then, for sufficiently large values of x, $ a^x &gt;&gt; b^x $; so $\sqrt[x]{a^x+b^x} \rightarrow \sqrt[x]{a^x} \rightarrow a $ when $x \rightarrow +\infty$.   Is that idea correct? How can I formalize it?
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$\lim_{x\to0}\dfrac{1}{x}-\lim_{x\to0}\dfrac{1}{x}$  Can we say answer is $0$ as $\lim_{x\to0}\dfrac{1}{x}-\dfrac{1}{x}=0$ or undefined because of $\infty-\infty$?
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The way I have understood, $0/0$ is undefined or indeterminate because, if $c=0/0$ then $c\cdot 0=0$, where $c$ can be any finite number including $0$ itself.  If we also observe a fraction $F=a/b$ where $a,b$ are positive real numbers,  the value of $F$ increases with the decrements of $b$. Being $0$ the least non-negative integer, if $b$ tends $0$ then $F$ tends to $\infty$ which is greater than all finite numbers.    I have also heard that no number is equal to infinity, a variable can tend to infinity. Any ratio is also a variable or is resolved to a number.  Now,   $$\tan(\pi/2) = \frac{\sin(\pi/2)}{\cos(\pi/2)}=\frac{1}{0},$$    is it undefined or infinity?  Also we know   $$\tan2x= \frac{2\tan x}{1-\tan^2x}$$  Is this formula valid for $x=\pi/2$?  Any rectification is more than welcome.  Some references:
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I've started studying calculus.  As part of studies I've encountered a question. How does one prove that $1 &gt; 0$?  I tried proving it by contradiction by saying that $1 &lt; 0$, but I can't seem to contradict this hypothesis.  Any help will be welcomed.
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Prove that if $a&gt;0$ then $1+a^9 \leq \frac{1}{a}+a^{10} $.   Using the fact that $a \gt 0$, multiply by $a$ on both sides and get everything to one side we have; $a^{11}-a^{10}-a+1 \geq 0$. By factoring $(a^{10}-1)(a-1) \geq 0 $.  I am not sure how to proceed any further.
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I am currently in high school and we are studying radicals. I had asked my math teacher why $\sqrt{ab}=\sqrt{a}\sqrt{b}$ (for all a,b>0) and he tries to prove it by arguing that $a^{1/2}*b^{1/2}=(ab)^{1/2}$ (an exponent law). However, I find this proof problematic since $x^{1/2}$ is simply defined as $\sqrt{x}$, so the reasoning is circular.   My view is that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ because once we square both sides we get $ab=ab$. Since we're obviously referring to the positive root, and the function $f(x) = \sqrt{x}$ is injective, it necessarily follows that the original expressions $\sqrt{ab}$ and $\sqrt{a}\sqrt{b}$ are equivalent because for injective functions it is not possible to map distinct elements in the domain to the same element in the range ($ab$). Hence they are equivalent expressions.   My question therefore is, is my proof valid and/or rigorous (I find it convincing but maybe it's wrong; I just want to be clear) and secondly was my teacher's proof correct?
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I have managed to use the root test and get as far as getting the limit to equal to $1$. But, I still need clarification on why the answer for the limit to this questions is $\alpha &lt; 1$. \begin{align} \sum _{n=0}^{\infty }\left( \frac {\alpha n} {n+1}\right) ^{n}\end{align} \begin{align} \lim _{n\rightarrow \infty }\left( \left( \frac {\alpha n} {n+1}\right) ^{n}\right) ^{1/n} =\left| \frac {\alpha n} {n+1}\right| =\alpha \lim _{n\rightarrow \infty }\left| \frac {n} {n+1}\right|=1.\end{align}
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Why is $ 1^{\infty} \not= \lim_{t \to \infty} 1^t = 1$?  It seems like they're saying the same thing. After all, $1$ to the power of any positive integer is equal to $1$.  I suspect that I'm misunderstanding some fundamental concept and would appreciate it if someone could explain it to me.  Thank you.