Letf\left ( 2 \right )= 4\: and \: {f}'\left ( 2 \right )= 4 then

\lim_{x\rightarrow 2}\frac{xf\left ( 2 \right )-2f\left ( x \right )}{x-2} equals

  • Option 1)

    2

  • Option 2)

    -2

  • Option 3)

    -4

  • Option 4)

    3

 

Answers (1)

As we learnt in 

L - Hospital Rule -

In \:the\:form\:of\:\:\;\frac{0}{0}\:\:and\:\:\frac{\infty }{\infty }\:\:\:we\:differentiate\:\:\frac{N^{r}}{D^{r}}\:\:separately.


\Rightarrow \lim_{x\rightarrow a}\:\:\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\:\:\frac{f'(x)}{g'(x)}

- wherein

\lim_{x\rightarrow a}\:\:\frac{\frac{d}{dx}\:f(x)}{\frac{d}{dx}\:g(x)}


Where \:\:f(x)\:\:and\:\:g(x)=0

 

 \lim_{x \to 2} \: \: \: \frac{xf(2)-2f(x)}{x-2}

\lim_{x \to 2} \: \: \: \frac{xf(2)-2{f}'(x)}{1}

        =f(2)-2{f}'(2) \: \: =4-2\times 4 \: \: =4-8 \: \: =-4


Option 1)

2

This option is incorrect

Option 2)

-2

This option is incorrect

Option 3)

-4

This option is correct

Option 4)

3

This option is incorrect

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