Proving a limit exists using Squeeze Theorem

Question

Let $g(x) = \ln(1+x)$ for $x>-1$. Use the inequality $\lim_{x \to 0}{x-\frac{x^2}{2}} \leq \lim_{x \to 0}{\ln(1+x)} \leq \lim_{x \to 0}x$ to prove that $g$ has a limit at $x = 0$.

For the left side of the inequality: Let $\epsilon>0$. Take $\delta = \min(\epsilon, \sqrt{2\epsilon})$. Then, we have that $|x| < \delta$. We note that, if $x$ is in the interval $(-1, 1)$, $|x-\frac{x^2}{2}| \leq |x| < \delta = \epsilon$. Now if $x$ is in the complement of that interval, we note that $|x-\frac{x^2}{2}| \leq |\frac{x^2}{2}|.$ Moreover, $|x|<\sqrt{2\epsilon} \implies |\frac{x^2}{2}| < \epsilon $. Thus, this function has a limit at x = 0

For the right side of the inequality: Let $\epsilon > 0$. Take $\delta = \epsilon$. We have that $|x| < \delta = \epsilon$. Thus the limit of this function exists at $x = 0$. By the squeeze thereom, $g$ has a limit at $x = 0$

Is my proof correct?

Answer 1

There is an easier way to do this using the Squeeze Theorem.

It seems you are trying to prove that $\lim_{x \to 0} \ln(1+x) = 0$ via the definition of a limit. But using the inequality that was given $$\lim_{x \to 0} x- \frac{x^2}{2} \leq \lim_{x \to 0} \ln(1+x) \leq \lim_{x \to 0} x$$

and the knowledge that $x- \frac{x^2}{2}$ and $x$ are continuous functions (since they are polynomial functions as Gabrielek points out above) we know that $\lim_{x \to 0} x- \frac{x^2}{2} = 0 - \frac{0^2}{2} = 0$ and $\lim_{x \to 0} x = 0$. Can you now see how to use the Squeeze Theorem and the above inequality to conclude that $\lim_{x \to 0} \ln(1+x) = 0$?

Answer 2

We assume as verified the continuity of $x-\frac{x^2}{2}$, $\ x$, $\ \text{log}(1+x)$ in a neighborhood of $0$.

In particular $x-\frac{x^2}{2}$, $x$ are both polynomials and their limits as $x$ approaches $0$ is of course $0$.

To conclude using the squeeze theorem we have to demonstrate that in a neighborhood of $0$ hold the following inequalities: $$ x-\frac{x^2}{2} \le \text{log}(1+x) \le x$$

Because $x \rightarrow 0$ , we can expand $\text{log}(1+x)$ in series: $$ \text{log}(1+x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n} = x -\frac{x^2}{2} + \frac{x^3}{3} + o(x^3)$$

And this concludes the left inequality. The right one follows as the other.

Answer 3

The Squeeze theorem says that

$$x-\frac{x^2}2\le\log(x+1)\le x\implies \lim_{x\to0}\left(x-\frac{x^2}2\right)\le\lim_{x\to0}\log(x+1)\le \lim_{x\to0}x,$$ and if the extreme limits are equal, the middle limit exists.

And this is just $$0\le\lim_{x\to0}\log(x+1)\le0.$$

Answer 4

Well the inequality $$\log(1+x)<x\,\,\forall x>-1\tag{1}$$ is a fundamental property of the logarithm function and alongwith the property $$\log(xy) =\log x+\log y\, \, \forall x, y>0\tag{2}$$ is sufficient for proving all the properties of logarithm function.

Replacing $x$ by $-x/(1+x)$ in $(1)$ we get $$\log\frac{1}{1+x}<-\frac{x}{1+x}$$ or $$\frac{x} {1+x}<\log(1+x)$$ Thus we have $$\frac{x} {1+x}<\log (1+x)<x\,\,\forall x>-1$$ You can now apply Squeeze theorem to reach desired conclusion.