Without L'Hospital's Rule, evaulate: $\displaystyle \lim_{x\to\infty} \frac { x \, ln( \, ln(x+1) \, ) } { ln(x) }$.
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For large x : $\displaystyle ln( ln(x+1) ) > 1 $ so $\displaystyle x ln( ln(x+1) ) > x $ so $\displaystyle \frac { x \, ln( \, ln(x+1) \, ) } { ln(x) } > \frac{x}{ln(x)} $ which clearly goes to infinity as x goes to infinity
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