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\infty
Spurningakeppni
Limits
5 vandamál svipuð og:
\lim_{ x \rightarrow 0 } \frac{1}{x^2}
Svipuð vandamál úr vefleit
Showing that the \lim_{x\to 0}\frac{1}{x^2} does not exist
https://math.stackexchange.com/q/1579837
Suppose that the limit exists and equals c\in\mathbb{R}. Then for e.g. \epsilon>1 some \delta>0 must exist with \left|x\right|<\delta\implies\left|\frac{1}{x^{2}}-c\right|<1. However, if we ...
Applying L'Hopital's rule to \lim\limits_{x \to 0}\frac{2}{x^2}
https://math.stackexchange.com/questions/502024/applying-lhopitals-rule-to-lim-limits-x-to-0-frac2x2
In order to use the 0/0 case of L'Hospital's rule, we require that both the numerator and the denominator tend to 0 at the appropriate point. The numerator does not tend to 0.
Is this piece-wise function continuous, and why?
https://math.stackexchange.com/questions/2411697/is-this-piece-wise-function-continuous-and-why
If we look at the behaviour as x approaches zero from the right, the function looks like this: \begin{matrix}x & f(x) = \frac{1}{x^2} \\ 1 & 1 \\ 0.1 & 100 \\ 0.01 & 10000 \\ 0.001 & 1000000 \\ 0.0001 & 100000000\end{matrix} ...
Manipulating \lim\limits_{x \to 0}{\frac{\sqrt{x+\sqrt{x}}}{x^n}}
https://math.stackexchange.com/questions/2177214/manipulating-lim-limits-x-to-0-frac-sqrtx-sqrtxxn
If \lim\limits_{x \to 0}{\frac{\sqrt{x+\sqrt{x}}}{x^n}} = c for some c\neq 0, then \lim\limits_{x \to 0}{\frac{x+\sqrt{x}}{x^{2n}}} =c^2. Now, let \sqrt{x}=t. We then wish to find n such ...
Limit of \frac{f'(x)}{g'(x)} & g'(x) \neq 0 in Hypotheses of L'Hospital's rule.
https://math.stackexchange.com/q/110408
When we write things like \lim_{x\to a}h(x) = \lim_{x\to a}H(x) we usually mean "if either limit exists, then they both do and they are equal; if either limit does not exist, then neither limit ...
How do we calculate the Right and Left Hand Limit of 1/x?
https://math.stackexchange.com/questions/762599/how-do-we-calculate-the-right-and-left-hand-limit-of-1-x
\mathbf{Definition} : \boxed{ \lim_{x \to a^+ } f(x) = \infty } means that for all \alpha > 0, there exists \delta > 0 such that if 0<x -a < \delta, then f(x) > \alpha \mathbf{Example} ...
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Svipuð vandamál
\lim_{ x \rightarrow 0 } 5
\lim_{ x \rightarrow 0 } 5x
\lim_{ x \rightarrow 0 } \frac{2}{x}
\lim_{ x \rightarrow 0 } \frac{1}{x^2}
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