Question

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


No spamming plz ​

Answer 1

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

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

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

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

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

↝ 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 :-

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

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

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

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

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