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1

At the time of my writing, there are 7 lengthy, beautifully written answers. They all, in one way or another, take as a launching point the definition of $e^x$ as an infinite series, or as a representation of a complex number in polar form on the complex plane. What an eight-grader might find puzzling is why one would approach e in this manner. There is a ...


1

[I see that Brian Tung has already posted a similar answer to this one. Although this is shorter, it's saying essentially the same thing, so I don't know if it will be easier to understand. Perhaps harder!] If: (1) you're familiar with the operation of differentiation of real-valued functions of real numbers; (2) you know that the function $x \mapsto e^x$ ...


2

As in my other answer, I won't be trying to prove anything to you -- power series, differential equations, etc. will be over your head for the moment. Instead I'll offer a different interpretation of the exponential function that hopefully you will find enlightening. Instead of seeing $e^{i\theta}$ as a number -- let's interpret it as an operator on ...


3

Write $$f(x)=x^{\sqrt x}$$ Then $$g(x)=\ln f(x)=\sqrt x\ln x=\frac{\ln x}{x^{-1/2}}$$ Now use l'Hopital to compute $$\lim_{x\to0^+}g(x)$$ Since $x\mapsto e^x$ is continuous, $$\lim_{x\to 0^+}f(x)=e^{\lim_{x\to0^+}g(x)}$$


2

Hint $$\sqrt x\ln x=\frac{\ln(x)}{\frac{1}{\sqrt x}}$$


0

Observe $$ e^{14} \;\; =\;\; \ln(e^e a) \;\; =\;\; \ln e^e +\ln a \;\; =\;\; e + \ln a $$ thus $\frac{1}{\ln a} = \frac{1}{e^{14} - e}$. Assuming NOTA stands for "none of the above" that should be your answer.



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