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Limits Continuity and Differentiability

Section: Mathematics

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Question : 4 of 61

Marks: +1, -0

If

f(x) = \begin{cases} \frac{(e^{a x}-1) \log(1+x)}{\sin^2 x}, & \text{if } x>0 \\ 2, & \text{if } x=0 \\ \frac{\cos 4x - \cos b x}{\tan^2 x}, & \text{if } x=0 \end{cases}

if continuous at

x=0

\sqrt{b^2 - a^2} =

[AP EAPCET 24 May 2025 S1]

4
5
3
7

Solution:

Given, function

f(x) = \begin{cases} \frac{(e^{a x}-1) \log(1+x)}{\sin^2 x}, & x>0 \\ \frac{\cos 4x - \cos b x}{\tan^2 x}, & x=0 \end{cases}

is contiuous at

x=0

Since,

f(x)

is continuous at

x=0

So,

\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = f(0)

and

f(0)=2

Now,

\lim\limits_{x \to 0^+} \frac{(e^{a x}-1) \log(1+x)}{\sin^2 x}

Applying L'hospital rule, twice, since it is an indeterminate form

\frac{0}{0}

.

So,

\lim\limits_{x \to 0^+} \frac{a e^{a x} \log(1+x) + \frac{E^{a x}-1}{1+x}}{2 \cos x \sin x}

This limit is again

\frac{0}{0}

form, so apply L'Hospital rule again.

\lim\limits_{x \to 0^+} \frac{a^2 e^{a x} \log(1+x) + \frac{a e^{a x}(1+x) - (e^{a x}-1)}{(1+x)^2} + \frac{a e^{a x}}{1+x}}{2 \cos^2 x - 2 \sin^2 x}

= \frac{0 + \frac{a(1)-(1-1)}{1^2} + \frac{a}{1}}{2(1)-0} = \frac{a+a}{2} = a

Since,

\lim\limits_{x \to 0^+} f(x) = f(0) \Rightarrow a=2

Now,

\lim\limits_{x \to 0^-} \frac{\cos 4x - \cos b x}{\tan^2 x}

= \lim\limits_{x \to 0^-} \frac{-2 \sin\left(\frac{4x+bx}{2}\right) \sin\left(\frac{4x-bx}{2}\right)}{\tan^2 x}

= \lim\limits_{x \to 0^-} \frac{-2 \sin\left(\frac{4x+bx}{2}\right) \sin\left(\frac{4x-bx}{2}\right) \cdot \cos^2 x}{\frac{\sin^2 x}{x^2} \cdot x^2}

\lim\limits_{x \to 0^-} \frac{ \frac{-2 \sin\left(\frac{4x+bx}{2}\right)}{\frac{4x+bx}{2}} \cdot \left(\frac{4x+bx}{2}\right) \cdot \frac{ \sin\left(\frac{4x-bx}{2}\right) }{ \frac{4x-bx}{2} } \cdot \left(\frac{4x-bx}{2}\right) \cdot \cos^2 x }{ \frac{\sin^2 x}{x^2} \cdot x^2 }

= \lim\limits_{x \to 0^-} \frac{-2 \cdot 1 \cdot \left(\frac{4x+bx}{2}\right) \cdot 1 \cdot \left(\frac{4x-bx}{2}\right) \cdot \cos^2 x}{1 \cdot x^2} \qquad \left[\because \lim\limits_{x \to 0} \frac{\sin x}{x}=1\right]

= \lim\limits_{x \to 0^-} \frac{-2 (16 x^2 - b^2 x^2)}{4 x^2} \cdot \cos^2 x

= \lim\limits_{x \to 0^-} \frac{b^2-16}{2} \cdot \cos^2 x = \frac{b^2-16}{2} \cdot 1 \qquad \left[\because \lim\limits_{x \to 0^-} \cos^2 x = 1\right]

= \frac{b^2-16}{2}

Also,

\lim\limits_{x \to 0^-} f(x) = f(0)

\Rightarrow \ \frac{b^2-16}{2} = 2 \Rightarrow b^2 = 4+16 = 20

Now,

\sqrt{b^2 - a^2} = \sqrt{20 - 2^2} = \sqrt{20-4} = \sqrt{16} = 4

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