# Calculating the limit of two unusually hard functions

I'm working my way through a single-variable analysis book right now, and the past limit problems have been no problem at all. But I got stuck at these two:

$$\lim_{x\to-\infty} \frac{x}{1 + \sqrt{x^2 + x + 1}}$$ \\ $$\lim_{x\to 0 } \frac{\ln(\cos x)}{x}$$

Direct insertion of the limit value results in undefined expressions, so the correct path must be to rewrite the functions right? My attempt at the first one went something like this:

$$\lim_{x\to-\infty} \frac{x}{1 + \sqrt{x^2 + x + 1}} = \lim_{x\to-\infty} \frac{x * (1 - \sqrt{x^2 + x + 1})}{1 - (x^2 + x + 1)}$$

But then I got completely stuck, and the second one I cannot even get anywhere with. What am I missing? What's the trick?

EDIT: I should add, I am not allowed to use L'hospitals rule.

• Are you familiar with L'Hopital rule? – user2566092 Sep 17 '15 at 20:38
• @user2566092 sorry I should have said, I am not allowed to use L'Hopitals rule – user262493 Sep 17 '15 at 20:39
• Can you use anything like $e^z \simeq 1 + z$ for $z$ small and $\cos z \simeq 1 - z^2/2$ for $z$ small? – user2566092 Sep 17 '15 at 20:42
• Divide numerator and denominator by $x^2$. – Aretino Sep 17 '15 at 20:46
• And for the second use $\cos x=\sqrt{1-\sin^2 x}$ combined with the well known limits of $\ln(1+t)/t$ and $\sin t/t$ as $t\to0$. – Aretino Sep 17 '15 at 20:48

Divide the numerator and denominator by $x$ to get

$\frac{1}{\frac{1+\sqrt{x^2+x+1}}{x}} = \frac{1}{\frac{1}{x} + \sqrt{1+\frac{1}{x} + \frac{1}{x^2}}}$

Now, taking the limit gives 1.

For the first one, observe that for large negative $x$, we have

$$\frac{x}{1+\sqrt{x^2+2x+1}} < \frac{x}{1+\sqrt{x^2+x+1}} < \frac{x}{1+\sqrt{x^2}}$$

which yields

$$\frac{x}{-x} < \frac{x}{1+\sqrt{x^2+x+1}} < \frac{x}{1-x}$$

or

$$-1 < \frac{x}{1+\sqrt{x^2+x+1}} < -1+\frac{1}{1-x}$$

Take the limit as $x \to -\infty$ and squeeze.

You don't have to use L'Hopital's rule here, it's not necessary. The denominator of the first one behaves as $|x|$ for large $x$ (sign does not matter), so the fraction is just $x/|x|=\text{sign}\,x=-1$ as $x\to-\infty$.

For the second one you have to Taylor expand the numerator for small $x$ where you use $\cos x\approx1-x^2/2$ and $\ln(1+x)\approx x$. Putting these together the numerator becomes $-x^2/2$ and the fraction is $-x/2$ for small $x$. The limit is thus zero.