GATE CSE 2008 | Question: 1

Question

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

• A. $1$
• B. $-1$
• C. $\infty$
• D. $-\infty$

Comments

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

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

 

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

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

 = 1

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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.

Answer 1

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

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

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

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:

$0\leq \dfrac{\sin x}{x} \leq 0 \text{ and } 0\leq \dfrac{\cos x}{x} \leq 0$
                         

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

Comments

yes, thanks

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So L hospital can be used when x tends to infinity, right???

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Yes L-hospital will work for all such x where we get 0/0 form or ∞ /∞  form .

Answer 2

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

Answer 3

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

Answer 4

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. 

Comments

LOL same wasted so much time on this for simple misconception

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There is no such rule that  L- Hospital Only works for x→ 0 and not For infinite.

L-hospital will work for all such x where we get 0/0 form or ∞ /∞  form .