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

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

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

= 1

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+14 votes

+43 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**

+1

$ = lim_{x \rightarrow \infty} \dfrac{Sin x}{x}$

$\text{Let x} = \dfrac{1}{t} \longrightarrow t = 0$

$ = lim_{t \rightarrow 0} \dfrac{Sin \dfrac{1}{t}}{\dfrac{1}{t}}$

$ = lim_{t \rightarrow 0} t*Sin \dfrac{1}{t}$

$ = 0$

Similarly

$ = lim_{t \rightarrow 0} t*Cos \dfrac{1}{t}$

$ = 0$

From above 2 forms of sinx and cosx, we get 1 as ans

$\text{Let x} = \dfrac{1}{t} \longrightarrow t = 0$

$ = lim_{t \rightarrow 0} \dfrac{Sin \dfrac{1}{t}}{\dfrac{1}{t}}$

$ = lim_{t \rightarrow 0} t*Sin \dfrac{1}{t}$

$ = 0$

Similarly

$ = lim_{t \rightarrow 0} t*Cos \dfrac{1}{t}$

$ = 0$

From above 2 forms of sinx and cosx, we get 1 as ans

+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

=> 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

+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

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

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