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mbrmbrg
- 496
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Why is [tex]1^\infty[/tex] an indeterminate form? If you keep multiplying 1 by itself doesn't the answer stay 1?
quasar987 said:Sure,
[tex]\lim_{x\rightarrow \infty} 1^x=1[/tex].
Maybe in some cases, the limit will converge to a value that is not 1? Can someone provide an exemple?
quasar987 said:Sure,
[tex]\lim_{x\rightarrow \infty} 1^x=1[/tex].
But consider a case like
[tex]\lim_{x\rightarrow \infty} (1+\frac{1}{x})^{e^{x^5}}[/tex]
1+1/x goes to 1, but does it so converges much slower than exp(x^5) goes to infinity! So if you think about it this way, maybe you see that it makes sense that sometimes a limit of the "form" [itex]1^{\infty}[/tex] will diverge, and some other times, it will behave and go to 1.
Maybe in some cases, the limit will converge to a value that is not 1? Can someone provide an exemple?
An indeterminate form is a mathematical expression that cannot be evaluated to a specific value using basic arithmetic operations. These forms typically involve the division of two quantities that approach infinity or zero.
When we take the limit of 1 raised to the power of infinity, the result is undefined because we do not know whether the exponent or the base is increasing faster. This uncertainty leads to an indeterminate form.
No, 1^∞ will always be an indeterminate form. However, we can use other mathematical techniques, such as L'Hospital's rule or logarithms, to evaluate the limit and determine its value.
Indeterminate forms, including 1^∞, are commonly used in calculus and other branches of mathematics. They can also be applied in physics, particularly in the study of limits and infinitesimals.
No, there are certain cases where the limit of 1^∞ can be evaluated to a specific value. For example, if the exponent is an odd integer, the limit will approach either 1 or negative 1 depending on the sign of the base. However, in general, 1^∞ is considered an indeterminate form.