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Definition: If, given any sequence $x_n \in D$ (where $D$ is a subset of the reals unbounded above), $\lim\limits_{x \to \infty} f(x)=L$ if, given any sequence $x_n \in D$ that diverges to $\infty$, $f(x_n) \to L$ (In the usual sense that, given any $\varepsilon>0$, there exists $N \in \Bbb{Z^+}$ such that for all $n \geq N, \vert x_n-L \vert <\varepsilon$)

Claim: $\lim\limits_{x \to \infty} f(x)=\lim_{x\rightarrow \infty}\frac{x+\sin(x)}{x+1}=1.$

Proof

Let $x_n$ be a sequence in $\mathbb{R}-\{1\}$ for which $x_n \to \infty.$ We must show that $f(x_n) \to 1.$ Let $K \in \mathbb{R}$ be given, and let $\varepsilon>0$ be given. Since $x_n \to \infty$, there exists $N\in \mathbb{Z^+}$ such that, for all $n \geq N, x_n>K.$

Now, for all $n \geq N:$ $\vert f(x_n)-1\vert=\vert \frac{x_n+\sin(x_n)}{x_n+1}\vert \leq \underbrace{\frac{\vert x_n \vert+\vert \sin(x_n) \vert}{\vert x_n+1 \vert}}_{\text{triangle inequality}}\leq\frac{\vert x_n \vert+1}{\vert x_n+1 \vert }<\underbrace{...}_{\text{what goes here?}}<\varepsilon$ Can someone (at least hint to) bridge the gap? Thanks

Samrat Mukhopadhyay
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beep-boop
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2 Answers2

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$$ \frac{x}{x+1} + \frac{\sin x}{x+1} $$

Given $\varepsilon>0$, find $N_1$ so large that whenver $x>N_1$, then $\dfrac{1}{x+1} <\dfrac\varepsilon2$.

Then $\left|\dfrac{\sin x}{x+1}\right|\le \left|\dfrac{1}{x+1}\right|<\dfrac\varepsilon2$.

Then finr $N_2$ so large that whenever $x>N_2$, then $\dfrac{x}{x+1}$ differs from $1$ by less than $\varepsilon/2$.

Then if $x>\max\{N_1,N_2\}$ then $$ \left|\frac{x+\sin x}{x+1} - 1\right| \le \left|\frac{x}{x+1} - 1 \right| + \left| \frac{\sin x}{x+1} \right| < \frac\varepsilon2+\frac\varepsilon2. $$

Michael Hardy
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  • But, according to this, $\vert f(x_n)-1\vert \leq \varepsilon,$ but it should be a strict inequality. – beep-boop May 06 '14 at 20:06
  • Got it. Are we assuming that, for the last last, $n \geq \max{N_1,N_2}$? – beep-boop May 06 '14 at 20:12
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    Typo: That last one was a strict inequality. Here's an exercise: Prove that if things like this work with weak inequalities ($\le$) then the also work with strict inequalities ($<$), provided $\varepsilon$ is strictly ($>$) positive. – Michael Hardy May 06 '14 at 20:12
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    @alexqwx : Correct. Very hasty typing today.... – Michael Hardy May 06 '14 at 20:13
  • One last thing- how does $\left| \frac{x}{x+1}\right|< \frac{\varepsilon}{2} \implies \frac{x}{x+1}<\frac{\varepsilon}{2}$, because the absolute value of a number is $\geq$ that number? – beep-boop May 06 '14 at 20:27
  • The inequality $\left|\dfrac{x}{x+1}\right|<\dfrac\varepsilon2$ occurs nowhere in my answer. – Michael Hardy May 06 '14 at 21:08
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Your proof has a tiny mistake that created the problem. According to your proof it should be $$|f(x_n)-1|=\left|\frac{x_n+\sin x_n}{x_n+1}-1\right|=\left|\frac{\sin x_n-1}{x_n+1}\right|<\frac{2}{x_n+1}$$by triangle inequality and hence goes to $0$ as $x_n\rightarrow \infty$. SO your proof is complete.

Samrat Mukhopadhyay
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  • And, using the triangle inequality on your last result (and using the fact that $x_n \to \infty$, how would I show that $ \vert f(x_n)-1 \vert < \varepsilon$? – beep-boop May 06 '14 at 20:02