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For the particular problem at hand, the method isn't as fast as the approach in the comments; but it's routine. $$\frac{dx}{\sqrt{C-x^2}}=\pm dt$$ $$\int{\frac{dx}{\sqrt{C-x^2}}}=\pm \int{dt}$$ $$\tan^{-1}\frac{x}{\sqrt{C-x^2}}=\pm t + K$$ $$\frac{x}{\sqrt{C-x^2}}=\tan{(\pm t + K)}\,.$$ Square both sides, rearrange, use the identity $1+\tan^2 ...
1) x^a × x^b = x^a+b; for x = 0 and a = 0, you would get 0^0 × 0^b = 0^b = 0, so we can't tell anything -- except confirm that 0^0 = 1 still works here! 2) x^{-a}=1/{x^a} -- so when a = 0 , x^{-0} = 1/x^0 = x^0 , which again does work for 0^0 = 1 ; 3) {x^a}^b = x^{a×b} , thus x^(1/n) is the n-th root -- and 1/n = 0 for no value of n , so ...
Since $0$ is the neutral element for the addition, we have that $$0x = (0 + 0)x$$ and because of distributivity we find that $$(0 + 0)x = 0x + 0x.$$ Hence we find that $$0x = 0x + 0x$$ so $0x$ also acts as the neutral element.
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$\begingroup$ You can't calculate exact value of sin(x)/x for x=$0$. When you say x tends to $0$, you're already taking an approximation.So, we have to calculate the limit here.Taylor series gives very accurate approximation of sin(x), so it can be used to calculate limit. $\endgroup$ –
The region to the left of 0. The type of graph will depend on whether is it a simple number line graph, or on a set of x-y- axes. On a number line graph, we would start with an open circle on 0, because 0 IS NOT included in the solution. Draw a line extending to the left, indicating that x can be any value to the left of 0. On ax-y- grid, we would have a vertical line to represent x = 0, but ...
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There is no x such that e^x = 0 The function e^x considered as a function of Real numbers has domain (-oo, oo) and range (0, oo). So it can only take strictly positive values. When we consider e^x as a function of Complex numbers, then we find it has domain CC and range CC "\\" { 0 }. That is 0 is the only value that e^x cannot take. Note that e^(x+yi) = e^x e^(yi) = e^x(cos y+i sin y) We have ...
When the right side is the zero vector, we know $\mathbf{x} = \mathbf{0}$ is a solution. Any other solutions (which we would call nontrivial) would signal that the matrix isn't invertible, since we would have more than one solution.
My question is: Show that $\lim_{x \rightarrow c} \frac{x^c-c^x}{x^x-c^c}$ exists and find its value. Because the limit is 0/0 I've tried using L'Hopital's rule, but every time I differentiate it I