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Limits Continuity and Differentiability question

2016 · Shift 1 · Q36

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JEE Advanced
Mathematics
Limits Continuity and Differentiability
2016 · Shift 1 · Q36

Limits Continuity and Differentiability question

2016 · Shift 1 · Q36

JEE AdvancedMathematicsLimits Continuity and DifferentiabilityNumerical+3 / −1

Let

\alpha

,

\beta\in

R be such that

\mathop {\lim }\limits_{x \to 0} {{{x^2}\sin (\beta x)} \over {\alpha x - \sin x}} = 1

. Then 6(

\alpha

+

\beta

) equals

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 7

We need to evaluate

\lim_{x\to 0} \frac{x^2\sin(\beta x)}{\alpha x-\sin x}=1.

We must find real numbers $\alpha,\beta$ for which this limit exists and equals $1$ .

Use the standard expansion near $x=0$ :

\sin x = x-\frac{x^3}{6}+O(x^5).

Also,

\sin(\beta x)=\beta x-\frac{(\beta x)^3}{6}+O(x^5).

So the numerator becomes

x^2\sin(\beta x)=x^2\left(\beta x-\frac{\beta^3x^3}{6}+O(x^5)\right) =\beta x^3+O(x^5).

Expand the denominator:

\alpha x-\sin x=\alpha x-\left(x-\frac{x^3}{6}+O(x^5)\right) =(\alpha-1)x+\frac{x^3}{6}+O(x^5).

For the limit to be finite and nonzero, numerator and denominator must have the same lowest power of $x$ .

Numerator starts with order $x^3$ .
Therefore denominator must also start with order $x^3$ .

Hence the coefficient of $x$ in the denominator must vanish:

\alpha-1=0 \implies \alpha=1.

Substitute $\alpha=1$ . Then denominator becomes

x-\sin x=\frac{x^3}{6}+O(x^5).

So

\lim_{x\to 0}\frac{x^2\sin(\beta x)}{x-\sin x} = \lim_{x\to 0}\frac{\beta x^3+O(x^5)}{\frac{x^3}{6}+O(x^5)} =\frac{\beta}{1/6}=6\beta.

Given that this limit equals $1$ ,

6\beta=1 \implies \beta=\frac16.

Now compute

6(\alpha+\beta)=6\left(1+\frac16\right)=6\cdot \frac76=7.

Therefore the required integer is

\boxed{7}.

Comparison with stored answer: Stored correct answer = $7$ , which matches our result.

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