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I feel that p>q because lt (x/sinx) = 1 and if p>q f(x) is continuous.But given answer is p>0.

Can someone explain how is b the answer ?

Random Walker
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  • Note that for $ p = 1 $ and $ q = 5 $, the function is not continuous around $ 0 $. So (B) cannot be right answer, can you check your answer key again? –  Mar 29 '20 at 14:55

2 Answers2

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Hint: To answer this question, you could look at the Taylor Series of $ \sin{x} $, which can be represented as -

$$ \sin{x} = x - \frac{x^3}{3!} + \frac{x^5}{5!} ........$$

Now, note that your function will be continuous at zero only if:

$$ \lim_{x \to 0} f(x) = f(0) = 0 $$

Now, for $ \lim_{x \to 0} f(x) $ to be equal to $0$, the numerator has to decrease faster than the denominator.

The highest power of $x$ in the numerator is $p$, while the highest power in the denominator can be found using the Taylor Series as mentioned above and it is $q$.

Therefore, $ \lim_{x \to 0} x^{p} < \lim_{x \to 0} x^{q} $, which corresponds to $p > q$ which is option A.

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Actually you have :

$$ \dfrac{x^p}{\sin(x)^q} \simeq \dfrac{x^p}{x^q}=x^{p-q}$$

which tend to zero if and only if $$ p>q $$ which is the right answer (a)

EDX
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