Here is the problem I have

$\lim \limits_{x \to -1} (x + 1)^2 sin (\frac{1}{x + 1})$

I approached it like this:

\begin{align} -1 \le sin(\displaystyle \frac{1}{x + 1}) \le 1 \\ -(x + 1) \le sin(1) \le (x + 1) \end{align}

I then go on to solve the limit by replacing $sin (\frac{1}{x + 1})$ with $-(x + 1)$ and $(x + 1)$ respectively.

Although this yields the correct answer, I am unsure if this is the correct way to solve this type of problem using the Squeeze theorem. Is this correct?


There appears to be a silly mistake which @mm-aops touched on, but I want to point out more directly.

If we have an inequality like so:

$$ a \le f\left( \frac{b}{c} \right) \le d $$ then we cannot conclude that $$ a \cdot c \le f(b) \le d \cdot c $$ If you want a direct counter example consider $f(x) = x^2$ and $a = 1, b = 4, c = 2, d = 5$ then we have surely $$ 2 \le 4 \le 5 \implies 2 \le f\left( \frac{4}{2} \right) \le 5 $$ but it is not true that $$ 4 \le f(4) \le 10 $$ since $f(4) = 16$.

Now with this said you started off correctly noting that $\sin(\cdot)$ is bounded above by $1$ and below by $-1$ i.e. $$ -1 \le \sin \left( \frac{1}{x+1} \right) \le 1 $$ Now since we're interested in the limit of $(x+1)^2 \sin \left( \frac{1}{x+1} \right)$ we should try to bound that. Do you see where to go from here?


it's not, you should've just written that $|\sin (\frac{1}{x+1})|$ is bounded by $1$ (which you did) and then say that $(x+1)^2 \rightarrow 0$, hence the product goes to zero as well, kinda like $$ - (x+1)^2 \leq (x+1)^2 \sin (\frac{1}{x+1}) \leq (x+1)^2$$ it isn't true that $\sin(\frac{1}{x+1}) = (x+1)\sin(1)$ which seems to be what you wrote

  • $\begingroup$ OK, that makes sense. I see where I fell off the cart $\endgroup$ – Leon Jun 3 '14 at 20:31

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