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

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

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

= 1

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Best answer

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

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

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

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,

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

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

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

answer = **option A**

$ = 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

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

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

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

Yes, thats what you do. Here it gives you 0/0 so you differentiate numerator and denominator which gives you :

lim(x->1) (2x-0)/(1-0) = 2

Which came from the fact that “1” is a constant so its derivate = 0.

at infinity, since we know sin and cos will give a finite value, we could simply ignore it, this is always not the case if some specific value of x is given, for it we would need to solve. But this is kind of intuitional, ofcourse the numerical methods are mathematical proofs that makes you aggree with the answer but intuitional is faster anytime.

lim(x->1) (2x-0)/(1-0) = 2

Which came from the fact that “1” is a constant so its derivate = 0.

at infinity, since we know sin and cos will give a finite value, we could simply ignore it, this is always not the case if some specific value of x is given, for it we would need to solve. But this is kind of intuitional, ofcourse the numerical methods are mathematical proofs that makes you aggree with the answer but intuitional is faster anytime.

Yes, thats what you do. Here it gives you 0/0 so you differentiate numerator and denominator

Then why didn’t you differentiate numerator and denominator for the given question ? I did the same what you have used in the answer and getting 0/0. What is the exact procedure ?

at infinity, since we know sin and cos will give a finite value, we could simply ignore it, this is always not the case if some specific value of x is given, for it we would need to solve.

in $\lim_{x \rightarrow \infty }(1-\cos x),$ cos gives a finite value, so can we ignore it ? what are those specific values of x for which we have to solve ?

But this is kind of intuitional, ofcourse the numerical methods are mathematical proofs that makes you aggree with the answer but intuitional is faster anytime.

I disagree with the answer because L’Hospital’s rule is not applicable here.

intuitive answer is fine but it should be justified with reasoning. Sometimes it works and sometimes doesn’t, so, it should be written when we have to ignore some function in what conditions.

L’Hopital’s rule is not only about 0/0 or $\infty/\infty$, we should also have to check whether $\lim_{x \rightarrow \infty }\frac{f'(x)}{g'(x)}$ exist or not. Since, here, $\lim_{x \rightarrow \infty }\frac{f'(x)}{g'(x)} =\; $$ \lim_{x \rightarrow \infty }\frac{(1-\cos x)}{(1- \sin x)}$ does not exist. So, L’Hopital’s rule is not applicable.

L’Hopital’s rule is not only about 0/0 or $\infty/\infty$, we should also have to check whether $\lim_{x \rightarrow \infty }\frac{f'(x)}{g'(x)}$ exist or not. Since, here, $\lim_{x \rightarrow \infty }\frac{f'(x)}{g'(x)} =\; $$ \lim_{x \rightarrow \infty }\frac{(1-\cos x)}{(1- \sin x)}$ does not exist. So, L’Hopital’s rule is not applicable.

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