in Calculus edited by
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$\displaystyle \lim_{x \to \infty}\frac{x-\sin x}{x+\cos x}$ equals

  1. $1$
  2. $-1$
  3. $\infty$
  4. $-\infty$
in Calculus edited by
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$\lim ( x - \sin x ) / (x + cos x )$

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

 

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

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

 = 1
4
One simple intuition is, $sin x$ and $cos x$ values are going to range between -1 and 1, so we can neglect $sin x$ and $cos x$ when $x\rightarrow \infty$, so the equation becomes $\lim_{x\rightarrow \infty }$$\frac{x}{x}$ , hence answer is 1.
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6 Answers

48 votes
 
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

edited by

8 Comments

$ = 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
2
what if we want to do it with L'hospital rule ? Please explain.
0
L-hospital applicable for $x\rightarrow 0$ , not for $x\rightarrow \infty$
0
I didn't knew that, thank you so much.
0
edited by
@srestha Can you please elaborate?
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@srestha

L’hsopital can be used infi/infi form also:

Here is an example involving ∞/∞:

Repeatedly apply L'Hôpital's rule until the exponent is zero (if n is an integer) or negative (if n is fractional) to conclude that the limit is zero.

https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule

3
yes, thanks
0
So L hospital can be used when x tends to infinity, right???
0
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
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
2 votes

When I give this full length test, I tried for L – Hospital Rule.

But it doesn’t work :(

After exam I find L- Hospital Only works for x→ 0 and not For infinite.

The simple solution for this is Just take x common from numerator and denominator. 

1 comment

LOL same wasted so much time on this for simple misconception
0
1 vote
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

5 Comments

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

if $\lim_{x \rightarrow 1} \frac{x^2 - 1}{x-1}$

Here, $x^2$ and $x$ also give finite values, So, here, we should also have to ignore, right ?

0
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.
0

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. 

0
edited by
You’re not getting it, I gave the intuitional answer, you better go the numerical way if you don’t get the intuition.
0
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.
2
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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