Without using differentiation, logarithmic function, rigorously, prove that $$e^x\ge x+1$$ for all real values of $x$.
Prove that $e^x\ge x+1$ for all real $x$
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$\begingroup$
analysis
inequality
exponential-function
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0Please avoid using display math (double \$\$) in titles. – 2012-12-06
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0Are you alowed to use MVT? – 2012-12-06
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6How do you define $e$ and/or $e^x$? What kind of answer is expected depends on the definition you are using. – 2012-12-06
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0Yes,I can use MVT, I want to prove it using only lim(1+x/n)n as n goest to inf – 2012-12-06
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0For positive x's I am done,but negative,I need help, have been trying it for hours... – 2012-12-06
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0Try using Bernoulli's inequality, see here http://en.wikipedia.org/wiki/Bernoulli%27s_inequality#Proof_For_Rational_Case – 2012-12-06
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0This is wrongly marked as duplicate since it from Dec 2012 and the other post is from 2013 – 2017-11-10