Limx- 1 - -ax - sinx-1 - ax- sinx-1 -1 -

Using L'Hopital's rule( 00 ): lim_x→ 1 [ ax + sin(x 1) + ax+ sin(x 1) 1 ] = lim_x→ 1 [ a + cos(x 1) #160; 1+ cos(x 1) ]= a+12 ...

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Standard XII

Mathematics

Derivative from First Principle

limx→ 1 [ -ax...

Question

${lim}_{x \to 1} [\frac{- a x + s i n (x - 1) + a}{x + s i n (x - 1) - 1}] =$

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Solution

Using $L^{'} H o p i t a l^{'} s$ rule( $\frac{0}{0}$ ):

${lim}_{x \to 1} [\frac{- a x + s i n (x - 1) + a}{x + s i n (x - 1) - 1}]$ = ${lim}_{x \to 1} [\frac{- a + c o s (x - 1)}{1 + c o s (x - 1)}] = \frac{- a + 1}{2}$

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limx→1[−ax+ sin(x−1)+ax+ sin(x−1)−1]=

Using L′Hopital′s rule(00):limx→1[−ax+ sin(x−1)+ax+ sin(x−1)−1]=limx→1[−a+ cos(x−1)1+ cos(x−1)]=−a+12

${lim}_{x \to 1} [\frac{- a x + s i n (x - 1) + a}{x + s i n (x - 1) - 1}] =$

Using $L^{'} H o p i t a l^{'} s$ rule( $\frac{0}{0}$ ):

${lim}_{x \to 1} [\frac{- a x + s i n (x - 1) + a}{x + s i n (x - 1) - 1}]$ = ${lim}_{x \to 1} [\frac{- a + c o s (x - 1)}{1 + c o s (x - 1)}] = \frac{- a + 1}{2}$