Brodie Beck

2022-09-05

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?

$$\underset{x\to \mathrm{\infty}}{lim}\frac{\sqrt{\mathrm{log}(x)+1}}{\mathrm{log}(\mathrm{log}(x))}=$$

I've been stuck with calculating the limit of the following problem for a while now. Can you help?

$$\underset{x\to \mathrm{\infty}}{lim}\frac{\sqrt{\mathrm{log}(x)+1}}{\mathrm{log}(\mathrm{log}(x))}=$$

Iademarco11

Beginner2022-09-06Added 10 answers

Substitute $u=\mathrm{log}(x)$. Then, since you get a $\frac{\mathrm{\infty}}{\mathrm{\infty}}$ expression, you can apply L'Hospital's rule:

$$\underset{x\to \mathrm{\infty}}{lim}\frac{\sqrt{\mathrm{log}(x)+1}}{\mathrm{log}(\mathrm{log}(x))}=\underset{u\to \mathrm{\infty}}{lim}\frac{\sqrt{u+1}}{\mathrm{log}(u)}=\underset{u\to \mathrm{\infty}}{lim}\frac{u}{2\sqrt{u+1}}=\mathrm{\infty}$$

$$\underset{x\to \mathrm{\infty}}{lim}\frac{\sqrt{\mathrm{log}(x)+1}}{\mathrm{log}(\mathrm{log}(x))}=\underset{u\to \mathrm{\infty}}{lim}\frac{\sqrt{u+1}}{\mathrm{log}(u)}=\underset{u\to \mathrm{\infty}}{lim}\frac{u}{2\sqrt{u+1}}=\mathrm{\infty}$$

reemisorgc

Beginner2022-09-07Added 1 answers

$$\underset{x\to \mathrm{\infty}}{lim}\frac{\sqrt{\mathrm{log}x+1}}{\mathrm{log}(\mathrm{log}x)}=\underset{x\to \mathrm{\infty}}{lim}\frac{\frac{1}{2x\sqrt{\mathrm{log}x+1}}}{\frac{1}{x\mathrm{log}x}}=\underset{x\to \mathrm{\infty}}{lim}\frac{\mathrm{log}x}{2\sqrt{\mathrm{log}x+1}}=\mathrm{\infty}$$

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