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

  1. $1$
  2. $-1$
  3. $\infty$
  4. $-\infty$
asked in Calculus by Veteran (59.4k points)
retagged by | 1.5k views
0
$\lim ( x - \sin x ) / (x + cos x )$

= $\lim ( 1 - \sin x/ x ) / (1 + cos x / x)$

 

when x -> inf , $\sin x / x and \cos x / x equals 0$

so $\lim ( 1 - 0 ) / (1 + 0 )$

 = 1

6 Answers

+30 votes
Best answer

$\lim_{x\rightarrow \infty} \dfrac{x-\sin x}{x+\cos x}$

$= \lim_{x\rightarrow \infty} \dfrac{x(1-\frac{\sin x}{x})}{x(1+\frac{\cos x}{x})}$

$= \lim_{x\rightarrow \infty} \dfrac{1-\frac{\sin x}{x}}{1+\frac{\cos x}{x}}$ 

now to calculate values of $\frac{\sin x}{x}$ and $\frac{\cos x}{x}$ we use Squeezing Theorem.

$-1\leq \sin x\leq {+1} \\ \dfrac{-1}{x}\leq \dfrac{\sin x}{x}\leq \dfrac{+1}{x}$ 

 

$-1\leq \cos x\leq {+1} \\ \dfrac{-1}{x}\leq \dfrac{\cos x}{x}\leq \dfrac{+1}{x}$ 

now as $x \rightarrow \infty$ we get $\frac{1}{x} \rightarrow 0$, this implies that:

$0\leq \dfrac{\sin x}{x} \leq 0 \\ 0\leq \dfrac{\cos x}{x} \leq 0$
                         ​​​​​​​

Hence, 
$\lim_{x\rightarrow \infty} \frac{x-\sin x}{x+\cos x} $

$= \lim_{x\rightarrow \infty} \frac{1-\frac{\sin x}{x}}{1+\frac{\cos x}{x}}$

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

answer = option A

answered by Boss (30.6k points)
edited by
+3 votes
lim x->inf (x-sinx)/(x+cosx)

=> lim x->inf (1- sinx/x)/(1+cosx/x)  since[       -1<=sinx<= +1   -1/x<= sinx/x<=+1/x at x->inf 0<= sinx/x<= 0 same for cosx, -1<=cosx<= +1 , -1/x<= cosx/x<=+1/x, at x->inf 0<= cosx/x<= 0 ]

 

 =>(1-0)/(1+0) = 1/1=1
answered by Boss (14.1k points)
+2 votes
answer of question is (a)

x->infinity ((x-sinx)/(x+cosx))  = we common out x from numerator and denominator will cancel eachother

so the right equation will be x->infinity ((1-(sinnx/x))/(1+(cosx/x)) as we know that for sine and cos domain will be all real value but its range will beb fixed lie between {-1 to 1} so sinx/x give {-1 to 1}/infinite = 0 similarly for cosx/x=0

value of limit will be {1-0}/{1+0}=1
answered by Active (4.9k points)
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 "
answered by Active (2.6k points)
–1 vote
1 is correct answer
answered by (161 points)
–3 votes

Answer should be ( A)

answered by (437 points)
0
explain it


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