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# 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

Note by Shailendra Garg
4 years, 5 months ago

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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. · 4 years, 5 months ago

but likewise $\mathop {\lim }\limits_{x \to 0} \left( {1 + x^2 } \right)^{\frac{1}{{x^4 }}} {\rm{ approaches}}{\kern 1pt} \;{\rm{to}}\;{\rm{infinite}}$ · 4 years, 5 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. · 4 years, 5 months ago

By continuity of log, exp, and product, we have $\lim_{x\rightarrow 0} (1+x^2)^{1/x^4} = \lim_{x\rightarrow 0} ((1+x^2)^{1/x^2})^{\lim_{x\rightarrow 0}1/x^2} = e^{\lim_{x\rightarrow 0}1/x^2} = \infty$ · 4 years, 5 months ago

Can you elaborate further? · 4 years, 5 months ago

@O B (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 · 4 years, 5 months ago