confusion on limits

we generally use \[ \mathop {\lim }\limits_{x \to 0} \left( {1 + x^2 } \right)^{\frac{1}{{x^2 }}} = e \] how ever if exponent of a number approaching to \[ 1^ +
\]approaches to infinite then resultant should approach to infinite

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

Can you elaborate further?

o b - 8 years, 3 months ago

(1+x^2) is greater than 1 and its exponent i.e 1/x^2 approaches to infinite , then this limit should approach to infinity, but it is 'e'. I know 'e' can be easily proved used logarithms or infinite series expansion. but the confusion again lies with my former statement

SHAILENDRA GARG - 8 years, 3 months ago

You're making assumptions about growth that aren't correct.

o b - 8 years, 3 months ago

No that is not always true. A trivial example would be 0^x. In this case note that the base, i.e. (1+x^2) tends to 1 as x tends to 0. So the resultant expression doesn't approach infinity.

Sambit Senapati - 8 years, 3 months ago

but likewise $\mathop {\lim }\limits_{x \to 0} \left( {1 + x^2 } \right)^{\frac{1}{{x^4 }}} {\rm{ approaches}}{\kern 1pt} \;{\rm{to}}\;{\rm{infinite}}$

SHAILENDRA GARG - 8 years, 3 months ago

As x becomes smaller and smaller the exponent approaches infinity while the base, i.e. 1+x^2 gets closer and closer to 1. So, the net result can't be guessed by us like that.

Sambit Senapati - 8 years, 3 months ago

By continuity of log, exp, and product, we have $\lim_{x\rightarrow 0} (1+x^2)^{1/x^4} = \lim_{x\rightarrow 0} ((1+x^2)^{1/x^2})^{\lim_{x\rightarrow 0}1/x^2} = e^{\lim_{x\rightarrow 0}1/x^2} = \infty$

Kai Chung Tam - 8 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$