Just cancel x x and notice sin ( m ) \sin(m) becomes m m

Calculus Level 1

True or False:

lim x 0 sin ( m x ) n x = m n \displaystyle \lim_{x\rightarrow 0} \frac{\sin(mx)}{nx} = \frac{m}{n}

If m , n m,n are non-zero reals then the above equation holds true.

False True

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2 solutions

Nihar Mahajan
Feb 2, 2016

Note that as x 0 x \rightarrow 0 , the expression sin x x \frac{\sin x}{x} becomes 0 0 \frac{0}{0} . Hence we can use L' Hopital's Rule : lim x 0 sin x x = lim x 0 cos x 1 = cos ( 0 ) = 1 \displaystyle\lim_{x\rightarrow 0} \frac{\sin x}{x} = \lim_{x\rightarrow 0} \frac{\cos x}{1} = \cos(0) = 1

Hence we have:

lim x 0 sin ( m x ) n x = lim x 0 m sin ( m x ) m n x = lim x 0 m n × lim x 0 sin ( m x ) m x = m n \lim_{x\rightarrow 0} \frac{\sin(mx)}{nx} = \lim_{x\rightarrow 0}\frac{m\sin(mx)}{mnx} = \lim_{x\rightarrow 0}\frac{m}{n} \times \lim_{x\rightarrow 0}\frac{\sin(mx)}{mx} = \frac{m}{n}

Hence , the result is indeed true!

Moderator note:

That big line equation is not true. What are you missing from the central terms?

No need for Mr. L' Hopital's rule since x \rightarrow 0 implies mx \rightarrow 0. Hence limit simplifies to: m n ( lim x 0 sin ( m x ) ( m x ) ) = m n \dfrac{m}{n} ( \lim_{x\rightarrow 0}\frac{\sin(mx)}{(mx)})=\frac{m}{n}

Rishabh Jain - 5 years, 4 months ago

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You must prove lim x 0 sin ( m x ) m x = 1 \displaystyle\lim_{x\rightarrow 0} \frac{\sin(mx)}{mx}=1

Nihar Mahajan - 5 years, 4 months ago

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1< sin x x \frac{\sin x}{x} <cosx and apply Squeeze Theorem lim x 0 1 < lim x 0 sin x x < lim x 0 cos x \small {\lim_{x\rightarrow 0} 1<\lim_{x\rightarrow 0}\frac{\sin x}{x}<\lim_{x\rightarrow 0} \cos x} lim x 0 sin x x = 1 \Rightarrow \lim_{x\rightarrow 0}\frac{\sin x}{x}=1

Rishabh Jain - 5 years, 4 months ago

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@Rishabh Jain Yes correct. For the same purpose I applied L' Hopital's (Cause I learnt it recently).

Nihar Mahajan - 5 years, 4 months ago

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@Nihar Mahajan No problem... (+1) for a Neatly written solution.. : D

Rishabh Jain - 5 years, 4 months ago

@Rishabh Jain 1 is not less than s i n x x \frac{sin x}{x} when x 0 \text{ when } x \rightarrow 0 ...

Guillermo Templado - 5 years, 3 months ago

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@Guillermo Templado That's why.. Since lim x 0 1 = lim x 0 cos x = 1 \small {\lim_{x\rightarrow 0} 1=\lim_{x\rightarrow 0} \cos x=1} Hence by Squeeze theorem we also claim lim x 0 sin x x = 1 \small {\lim_{x\rightarrow 0}\frac{\sin x}{x}=1} .

Rishabh Jain - 5 years, 3 months ago

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@Rishabh Jain ok, look at this

Guillermo Templado - 5 years, 3 months ago

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@Guillermo Templado sin x < x < tan x when x > 0 \sin x < x < \tan x \quad \text { when } x >0 and x belongs to neighborhood of 0. Now , we are going to divide for sin x \sin x . Then, sin x sin x < x sin x < 1 cos x \frac{\sin x}{\sin x} < \frac{x}{\sin x} < \frac{1}{\cos x} and the we can apply squezze theorem when x 0 + x \to 0^{+} .(I can give you another arguments when x < 0) sin x \sin x is a vertical segment and x is an angle (we are talking in radians at the circrcle of radius 1 (geometry)) for that sin x < x when x > 0 \sin x < x \quad \text{ when } x > 0 . I can give you one analytical demonstration of this too, if you desire it

Guillermo Templado - 5 years, 3 months ago

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@Guillermo Templado Sure.. Please go ahead.. :-)

Rishabh Jain - 5 years, 3 months ago

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@Rishabh Jain Ok, take f ( x ) = x sin x f(x)= x - \sin x this function fullfies f ( 0 ) = 0 0 = s i n 0 f(0) = 0 \Rightarrow 0 = sin 0 and f ( x ) = 1 cos x > 0 when x ( 0 , π 4 ) f ' (x ) = 1 - \cos x > 0 \text{ when } x \in (0, \frac{\pi}{4}) so this function is strictly increasing in this interval , and this means x > sin x when x ( 0 , π 4 ) x > \sin x \text{ when } x \in (0, \frac{\pi}{4} ) ,you can apply the mean value theorem (Lagrange theorem) or also take Taylor polynomyal of sin x = x x 3 3 ! \sin x = x - \frac{x^3}{3!} ... graph x - sinx

When x < 0, then we have tan x < x < sin x \tan x < x < \sin x in a neighborhood of 0 (this time x < 0) and we can also apply the squeeze theorem when x 0 x \to 0^{-}

Guillermo Templado - 5 years, 3 months ago

Just use definition of derivative of Sin function at 0 witch lead to Cos(0) = 1 .

Sam Lee - 5 years, 4 months ago

In response to Challenge Master: Lol , I was being silly. I have corrected it now.

Nihar Mahajan - 5 years, 3 months ago
. .
Mar 15, 2021

It is always true. \text { It is always true. }

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