{θ →0}{lim}1-cosmθ /1-cosnθ =?

Correct Answer is : m2n2{m²}{n²}n2m2

Correct Answer is : m2n2{m²}{n²}n2m2

Test Index

AP DSC School Assistant Math Model Test 1

Section: Subject Specific

Question : 67 of 160

Marks: +1, -0

\underset{\theta →0}{lim}\frac{1-cosm\theta }{1-cosn\theta }=?

Solution:

Concept:

Use the trigonometric identity

1-\cos x = 2\sin^{2}\frac{x}{2}

and the standard limit

\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1

Explanation:

We need to evaluate

\displaystyle\lim_{\theta\to0}\frac{1-\cos(m\theta)}{1-\cos(n\theta)}

Replace each cosine term using the identity:

1-\cos(m\theta)=2\sin^{2}\frac{m\theta}{2}

and

1-\cos(n\theta)=2\sin^{2}\frac{n\theta}{2}

Thus the limit becomes:

\displaystyle\lim_{\theta\to0}\frac{2\sin^{2}\frac{m\theta}{2}}{2\sin^{2}\frac{n\theta}{2}}=\lim_{\theta\to0}\frac{\sin^{2}\frac{m\theta}{2}}{\sin^{2}\frac{n\theta}{2}}

Now express each sine term in the form

\frac{\sin(k\theta)}{k\theta}\cdot k\theta

\displaystyle\frac{\sin^{2}\frac{m\theta}{2}}{\sin^{2}\frac{n\theta}{2}}=\frac{\left(\frac{\sin\frac{m\theta}{2}}{\frac{m\theta}{2}}\right)^{2}\cdot\left(\frac{m\theta}{2}\right)^{2}}{\left(\frac{\sin\frac{n\theta}{2}}{\frac{n\theta}{2}}\right)^{2}\cdot\left(\frac{n\theta}{2}\right)^{2}}

\theta\to0

, each

\frac{\sin(\text{small})}{\text{small}}\to1

. Hence the limit simplifies to:

\displaystyle\frac{\left(\frac{m\theta}{2}\right)^{2}}{\left(\frac{n\theta}{2}\right)^{2}}=\frac{m^{2}\theta^{2}/4}{n^{2}\theta^{2}/4}=\frac{m^{2}}{n^{2}}

Answer:

The correct option is D:

\displaystyle\frac{m^{2}}{n^{2}}

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