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I'm confused how to prove that $\lim_{x\rightarrow 1} \sin (x^2 - 1) / (x^2 - 1) = \lim_{x\rightarrow 0} \sin(x)/x$.

Assuming $\lim_{x\rightarrow 1} \sin (x^2 - 1) / (x^2 - 1) = l$, then there's $\delta > 0$ s.t. for all $x$, if $0 < |x - 1| < \delta$ implies that $ | \sin (x^2 - 1) / (x^2 - 1) - l| < \epsilon $ for all $\epsilon > 0$.

My approach: if we take $|x| < \delta$, then $| (x + 1) - 1| < \delta$ this implies that $| \sin (x^2 - 2x) / (x^2 - 2x) - l| < \epsilon$, which probably isn't the route I need to take.

Intuitively it makes sense, since $x^2 - 1$ approaches $0$, then we should be able to replace it with its limit.

Gary
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Flight Booker
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  • $~~~~$ $x^2-1 = y$ – Enrico M. Sep 20 '20 at 19:14
  • https://en.wikipedia.org/wiki/Limit_of_a_function#Limits_of_compositions_of_functions – Gary Sep 20 '20 at 19:15
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    One way to proceed "from scratch" is to follow the usual argument that a composition of continuous functions is continuous. Namely, to see $f \circ g$ is continuous at $x_0$ if $g$ is continuous at $x_0$ and $f$ is continuous at $g(x_0)$, proceed as follows. Let $\epsilon > 0$, find $\delta_1$ such that if $|y-g(x_0)|<\delta_1$ then $|f(y)-f(g(x_0))|<\epsilon$, then find $\delta_2$ such that if $|x-x_0|<\delta_2$ then $|g(x)-g(x_0)|<\delta_1$. Then $f \circ g$ has the function $\epsilon \mapsto \delta_2$ as a modulus of continuity. – Ian Sep 20 '20 at 19:16
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    (Cont.) So in your setting, notice that $|x^2-1|<\epsilon$ if $|x-1|<\min { 1,\epsilon/4 }$, so if you have a valid $\delta_1$ for the $\frac{\sin(x)}{x}$ proof then you can just set $\delta_2=\min { 1,\delta_1/4 }$ to finish. – Ian Sep 20 '20 at 19:16

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This is for the theorem of composition since $f(x)=x^2-1 \to 0$ and $x^2-1 \neq 0$ for $x\neq 1$ the following holds

$$\lim_{x\rightarrow 1} \frac{\sin (f(x)) }{ f(x)} = \lim_{x\rightarrow 0} \frac{\sin(x)}x=1$$

Refer also to the related

user
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