If we apply L'Hopital's rule we get: which tends to infinity, so the limit is . When dealing with ratios such as 1/0, 0/0, or infinity in any form, you will most likely need to use a further theorem, such as L’Hopital’s, to solve for a limit. On the contrary, those limits tell you that the limit of the entire quotient is $0$. Indeterminate form of infinity over 0? ∞into 0 1/∞ or into ∞ 1/0, for example one can write lim x→∞xe −x as lim x→∞x/e xor as lim x→∞e −x/(1/x). After substitution in a limit you can arrive at an indeterminate form like infinity * 0. Outside of the context of limits infinity * 0 would just be undefined for the reasons people have stated, multiplication is defined for numbers and infinity is not really a number. Such forms are: 1^∞, ∞^0, 0^0, 0/0, ±∞/±∞, 0*∞, ∞ - ∞ For 1^∞, you would think that the limit evaluates to 1 everytime this occurs. This may be easier to see if you rewrite to $$\lim_{x\to\infty} f(x)\frac1{h(x)}$$ where $\lim_{x\to\infty} f(x) = 0$ and $\lim_{x\to\infty} \frac1{h(x)}=0$, and the product of two functions that both have limit $0$ surely also has limit $0$. Substitute the value of x to the above equation, we have. We are assuming ∞ ∞ \frac{\infty}{\infty} ∞ ∞ is defined, which has been disproven using a similar technique used in the problem. Indeterminate forms hover over the calculus no matter where you turn. Since the answer is ∞∙0 which is also another type of Indeterminate Form, it is not accepted in Mathematics as a final answer. As x approaches ∞, both x and lnx approach infinity, so this is an indeterminate limit of type ∞/ ∞. Solution: Consider the given equation. Viewed 4k times 1 $\begingroup$ I know that indeterminate forms exist in limits, such as $\frac{0}{0}$, $\frac{\infty}{\infty}$, $0^0$, $\infty^0$, $1^\infty$. 0/∞ is not an indeterminate form. 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