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Here $f(x) = x + \cos x$ and $g(x)= x- \cos x$

We can't apply L'Hopital's Rule since $\forall c > 0\quad \exists x_c \in (c, + \infty): g'(x) = 1 + \sin x = 0$.

But I don't know how to proceed.

My attempt:

Because $\cos x$ is bounded I was taking an arbitrary number $A >0$ that bounds $\cos x$ from above.

From there I have

$$\lim_{x\to \infty}\frac{x+\cos x}{x - \cos x} = \lim_{x\to \infty}\frac{x+A}{x - A} =1$$

But the answer is $0$. What am I doing wrong?

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    $\begingroup$ The answer is obviously $1$, you can plot the function with Wolfram Alpha to have a numerical proof. Algebraically you can divide the numerator and the denominator by $x$ to make it obvious. wolframalpha.com/input/… $\endgroup$ Commented Nov 9, 2016 at 12:13
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    $\begingroup$ Divide top and bottom by $x$. $\endgroup$ Commented Nov 9, 2016 at 12:14
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    $\begingroup$ Why do you say "But the answer is 0" ? $\endgroup$
    – Henry
    Commented Nov 9, 2016 at 12:14
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    $\begingroup$ "What am I doing wrong?" Nothing; you have the correct answer. $0$ is incorrect. $\endgroup$ Commented Nov 9, 2016 at 12:23

3 Answers 3

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The easiest way is divide numerator and denominator by x so you have

$$\lim_{x\to \infty}\frac{x+\cos x}{x - \cos x} = \lim_{x\to \infty}\frac{1+\frac{\cos{x}}{x}}{1 - \frac{\cos{x}}{x}} = \frac{1+0}{1-0} = 1$$

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For $x \gt 1$, you have $\dfrac{x+\cos x}{x - \cos x}=1 + \dfrac{2\cos x}{x - \cos x}$

Since $0 \le | \cos x | \le 1$ you have $-2 \le 2 \cos x \le 2$ and $x-\cos x \ge x-1$

meaning $1 - \dfrac{2}{x - 1} \le \dfrac{x+\cos x}{x - \cos x} \le 1 + \dfrac{2}{x - 1}$ and thus a limit of $1$ as $x$ increases without limit

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$\frac{x+\cos x}{x - \cos x}=\frac{1+\frac{\cos x}{x}}{1 - \frac{\cos x}{x}} \to 1$ for $x \to \infty$, since $0 \le \frac{|\cos x|}{x}\le \frac{1}{x}$ for $x>0$

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