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I have some function.

$$ \lim_{t\to3} \ \sin{\left(\frac{1}{t-3}\right)} \ e^t \ (t-3)^2 $$

I want to evaluate it using the Squeeze theorem but I do not know what it means to do this. I know that my limit is $0$ and I know that the squeeze theorem says if $f(x) \le g(x) \le h(x)$ then the limit of $g(x)$ must be the same as that of $f(x)$ and $h(x)$. However I do not understand how to implement this.

I would like to know how this works and what is happening with steps.

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Hint: $ -1 \le \sin{\left(\frac{1}{t-3}\right)} \le +1$

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  • $\begingroup$ Isn't $\sin\left(\frac{1}{t-3}\right)$ undefined when $t = 3$? $\endgroup$ – Computer Jun 8 '17 at 4:56
  • $\begingroup$ Yes, but for evaluating limit your function need to be defined on a deleted neighborhood of point . so you remove $t=3$ from real line then your function is well-defined @Computer $\endgroup$ – Red shoes Jun 8 '17 at 5:00
  • $\begingroup$ So what are you saying here? I am actually just looking at $\sin\left(\frac{1}{-3}\right)$? $\endgroup$ – Computer Jun 8 '17 at 5:29
  • $\begingroup$ What he's saying is that even without knowing the value of the sine function, you end up with a function $f(t)$ such that $-e^t(t-3)^2\leq f(t)\leq e^t(t-3)^2$, so your limit is squeezed between the limit of these two functions. $\endgroup$ – User123456789 Jun 8 '17 at 9:06
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You can ignore the factor $e^x$, since $e^x$ is independent of $t$. Therefore we only have to look at $\sin{\left(\frac{1}{t-3}\right)} \ (t-3)^2$. For $t \in (2,4) \setminus \{3\}$ we have

$0 \le |\sin{\left(\frac{1}{t-3}\right)} \ (t-3)^2| \le (t-3)^2$.

This gives $\sin{\left(\frac{1}{t-3}\right)} \ (t-3)^2 \to 0$ for $t \to 3$.

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  • $\begingroup$ Where do you get the interval (2,4)? $e^x$ is actually $e^t$... sorry. What is $\{3}$? $\endgroup$ – Computer Jun 8 '17 at 7:58

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