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

Calculus Level 1

True or False:

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

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

False True

2 solutions

Nihar Mahajan
Feb 2, 2016

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

Hence we have:

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

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: $\dfrac{m}{n} ( \lim_{x\rightarrow 0}\frac{\sin(mx)}{(mx)})=\frac{m}{n}$

Rishabh Jain - 5 years, 4 months ago

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

Nihar Mahajan - 5 years, 4 months ago

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

Rishabh Jain - 5 years, 4 months ago

@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

@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 $\frac{sin x}{x}$ $\text{ when } x \rightarrow 0$ ...

Guillermo Templado - 5 years, 3 months ago

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

Rishabh Jain - 5 years, 3 months ago

@Rishabh Jain – ok, look at this

Guillermo Templado - 5 years, 3 months ago

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

Guillermo Templado - 5 years, 3 months ago

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

Rishabh Jain - 5 years, 3 months ago

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

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

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

$\text { It is always true. }$

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

2 solutions

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