Math, asked by SweetLily, 1 month ago

f(2) = 4 , f'(2) = 1, then
 \sf{lim_{x \to2} \frac{xf(2)-2f(x)}{x-2} } =  ?
a) 1
b) 2
c) 3
d) -2


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Answers

Answered by mathdude500
3

\begin{gathered}\begin{gathered}\bf \:Given-\begin{cases} &\sf{f(2) = 4} \\ &\sf{f'(2) = 1} \end{cases}\end{gathered}\end{gathered}

\begin{gathered}\begin{gathered}\bf \: To \: Find - \begin{cases} &\sf{\tt \: \displaystyle\lim_{x\to2} \sf \: \dfrac{xf(2) - 2f(x)}{x - 2} }\end{cases}\end{gathered}\end{gathered}

\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}

 \boxed{ \bf \: \displaystyle\lim_{x\to \: a} \bf \: \dfrac{f(x) - f(a)}{x - a} = f'(a)}

\large\underline{\sf{Solution-}}

↝ Given that

  • ↝ f(2) = 4

and

  • ↝ f'(2) = 1

↝ Consider,

\rm :\longmapsto\:\displaystyle\lim_{x\to2}\sf \: \dfrac{xf(2) - 2f(x)}{x - 2}

  • ↝On substituting directly x = 2, we get indeterminant form.

So,

  • ↝Simplify this limit, we add and subtract 2f(2) in numerator,

  • ↝So above expression can be rewritten as

 \:  \:  \:  =  \:  \: \:  \displaystyle\lim_{x\to2}\sf \: \dfrac{xf(2) - 2f(2) + 2f(2) - 2f(x)}{x - 2}

 \: \: \: =  \: \:   \: \displaystyle\lim_{x\to2}\sf \: \dfrac{f(2)\bigg(x - 2 \bigg)  - 2\bigg(f(x) - f(2) \bigg) }{x - 2}

 \sf \:  \:  \:  \:  =  \:  \:f(2) \displaystyle\lim_{x\to2} \: \dfrac{ \cancel{x - 2}}{ \cancel{x - 2}}  \:  -  \: 2 \: \displaystyle\lim_{x\to2} \: \dfrac{f(x) - f(2)}{x - 2}

 \sf \:  \:  \:  \:  =  \:  \: f(2) \:  -  \: 2 \: f'(2)

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \boxed{ \bf \because \: \displaystyle\lim_{x\to2} \bf \: \dfrac{f(x) - f(2)}{x - 2}  = f'(2)}

 \sf \:  \:  \:  \:  =  \:  \: 4 \:  -  \: 2 \:  \times  \: 1

 \sf \:  \:  \:  \:  =  \:  \: 4 - 2

 \sf \:  \:  \:  \:  =  \:  \: 2

\overbrace{ \underline { \boxed { \rm \therefore \: \displaystyle\lim_{x \: \to \: 2} \: \bf \: \dfrac{x \: f(2)  \: -  \: 2 \: f(x)}{x \:  -  \: 2} \:  =  \: 2}}}

\large{\boxed{\boxed{\bf{Option \:  (b) \:  is \:  correct}}}}

Additional Information :-

\begin{gathered}(1)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{sin \: x}{x} \:  =  \: 1 }}}}}} \\ \end{gathered}

\begin{gathered}(2)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{tan \: x}{x} \:  =  \: 1 }}}}}} \\ \end{gathered}

\begin{gathered}(3)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{ log(1 \:  +  \: x) }{x} \:  =  \: 1 }}}}}} \\ \end{gathered}

\begin{gathered}(4)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{ {e}^{x} \:  -  \: 1 }{x} \:  =  \: 1 }}}}}} \\ \end{gathered}

\begin{gathered}(5)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{ {a}^{x} \:  -  \: 1 }{x} \:  =  \:  log(a)  }}}}}} \\ \end{gathered}

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