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  • Help me please, - Limit , continuity and differentiability - JEE Main

Let k be a non - zero  real  number. If

  f(x=) \left \{ \right.\frac{(e^{x}-1)^{2}}{\sin\left ( \frac{x}{k} \right )\log \left ( 1+\frac{x}{4} \right )}, x\neq 0

                                        12                  ,x= 0

is a continuous function, then the value of k is :

 

  • Option 1)

    1

  • Option 2)

    2

  • Option 3)

    3

  • Option 4)

    4

 

Answers (1)

As we learnt in 

Evaluation of Exponential Limits -

\lim_{x\rightarrow 0}\:\:\:\frac{e^{x}-1}{x}=1

\lim_{x\rightarrow 0}\:\:\:\frac{a^{x}-1}{x}=log_{e}{a}

\lim_{x\rightarrow 0}\:\:\:\frac{e^{Kx}-1}{x}\neq 1

\therefore \:\:\:\lim_{x\rightarrow 0}\:\:\:\frac{e^{Kx}-1}{x}\times \frac{K}{K}

\therefore \:\:\:K\lim_{x\rightarrow 0}\:\:\:\frac{e^{Kx}-1}{Kx}=K\times 1=K

- wherein

\lim_{x\rightarrow 0}\:\:\:\frac{e^{x}-1}{x}

x\:must\:be\:same

 

 f(x)=\left\{\begin{matrix} \frac{(e^{x}-1)^{2}}{sin\left(\frac{x}{k} \right )log\left(1+\frac{x}{4} \right )} \, \, \, \, \, \, \, \, \, x\neq0 \\ 12\, \, \, \, \, \, \,\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, x=0 \end{matrix}\right.

 

 

\therefore \lim_{x\rightarrow \frac{0^{+}}{0^{-}}}\frac{\left (e^{x}-1 \right )^{2}}{\sin\left ( \frac{x}{k} \right )log(1+\frac{x}{4}) }

\therefore \lim_{x\rightarrow \frac{0^{+}}{0^{-}}}\frac{\frac{\left ( e^{x}-1 \right )^{2}}{x^{2}}}{\frac{sin\left ( \frac{n}{k} \right )log(1+\frac{r}{4})}{\frac{x}{k} \times k\, \, \, \, \, \frac{x} {4}\times 4}}

\lim_{x\to0}\frac{\left(\frac{e^{x}-1}{x} \right )^{2}}{\left(\frac{sin\frac{x}{k}}{\frac{x}{k}} \right )\frac{log\left(1+\frac{x}{4} \right )}{\frac{x}{4}}}\times 4k

\therefore \frac{1\times 4k}{1\times 1}=4k

\therefore 4k=12

\therefore k=3


Option 1)

1

Incorrect option    

Option 2)

2

Incorrect option    

Option 3)

3

Correct option    

Option 4)

4

Incorrect option    

Posted by

Sabhrant Ambastha

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