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$\displaystyle \lim_{x \to \infty}\frac{x-\sin x}{x+\cos x}$ equals

  1. $1$
  2. $-1$
  3. $\infty$
  4. $-\infty$
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9 Answers

1 votes
1 votes
sin(∞) = between -1 to +1 = finite

cos(∞) = between -1 to +1 = finite

So, if you put ∞;

 

(∞-finite) / (∞ – finite) = (∞/∞) form

Thus you can apply L-Hoptials’ Rule :

1/1 = 1 (Ignoring sinx and cosx because I know thats a finite value eventually, so differentiating them is not really going to help me with limit)

Thus, answer = 1
1 votes
1 votes
Form :- ($\infty$/→$\infty$) => indeterminate form.

$\lim_{x→\infty}$ [sin(x) / x] = 0, because the value of sin(x) is between [-1,1]. Similarly, $\lim_{x→\infty}$ [cos(x) / x] = 0, because the value of cos(x) is also between [-1,1]. (finite number/$\infty$ = 0) (Think intuitively)

$\lim_{x→\infty}$ (x – sin(x)) / (x + cos(x))
= $\lim_{x→\infty}$ (1 – sin(x)/x) / (1 + cos(x)/x)
= $\lim_{x→\infty}$ (1 – 0) / (1 + 0)
= 1
0 votes
0 votes
limx→∞ x−sinx / x+cosx

on dividing with X , we get (1 - sinx/ x ) / ( 1 - cosx/x )

as we know sin x and cos x values are always constant between -1 to + 1 , so constant / infinity = 0

so on applying limits,     = limx→∞(1 - sinx/ x ) / ( 1 - cosx/x )

                                   = 1- 0 / 1-0 = 1                                                    " correct answer ( A) = 1 "
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