How do I calculate the limit of this log function? I've been stuck with calculating the limit of the following problem for a while now. Can you help?
$$ \lim_{x\to \infty} \frac{\sqrt{\log(x) + 1}}{\log(\log(x))} = $$
 A: Substitute $u = \log(x)$. Then, since you get a $\infty \over \infty$ expression, you can apply L'Hospital's rule:
$$\lim_{x \to \infty} \frac{\sqrt{\log(x)+1}}{\log(\log(x))} = \lim_{u \to \infty} \frac{\sqrt{u+1}}{\log(u)} = \lim_{u\to \infty} {u\over 2  \sqrt{u+1}} = \infty$$
A: $$
\lim_{x\to\infty}\frac{\sqrt{\log x+1}}{\log(\log x)}=\lim_{x\to\infty}\frac{\frac1{2x\sqrt{\log x+1}}}{\frac1{x\log x}}=\lim_{x\to\infty}\frac{\log x}{2\sqrt{\log x+1}}=\infty
$$
using L'Hôpital's rule.
A: Solution is +oo.
Say that lnx=t. x->00 therefore t->oo
now you hav lim sqrt(t+1)/lnt when t->oo.
You can use L'Hopital's rule and you get
lim t/2*sqrt(t+1) when t->00
lim t/2*sqrt(t)*sqrt(1+1/t) t->oo
lim sqrt(t)/2*sqrt(1+1/t)=oo when t->oo
A: $$\lim_{x\to\infty}\frac{\sqrt{\log x+1}}{\log\log x}\stackrel{\text{l'Hospital}}=\lim_{x\to\infty}\frac{\frac1{2x\sqrt{\log x+1}}}{\frac1{x\log x}}=\lim_{x\to\infty}\frac12\frac{\log x}{\sqrt{\log x+1}}\stackrel{\text{l'H}}=$$
$$=\lim_{x\to\infty}\frac12\sqrt{\log x+1}=\infty$$
A: First note $\sqrt{\log x+1}\sim_\infty \sqrt\log x$, hence
$$\frac{\sqrt{\log x+1}}{\log\log x}\sim_\infty \frac{\sqrt\log x}{\log\log x}$$
Now set $u=\log x$; we have
$$\frac{\sqrt\log x}{\log\log x}=\frac{\sqrt u}{\log u}\xrightarrow[u\to \infty]{}+\infty. $$
A: You also can define $\log(\log(x))=y$ which makes $\log(x)=e^y$ $$\frac{\sqrt{\log(x) + 1}}{\log(\log(x))}=\frac{\sqrt{e^y+1}}y=\frac{e^{y/2}}y \sqrt{1+\frac 1 {e^y}}\simeq \frac{e^{y/2}}y  $$ and use L'Hospital rule or just use the fact that the exponential grows much faster than its argument.
