Question

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and answer of this Q is10​

Answer 1

Question :–

• Find the value of given limit –

$\\ \implies\tt \lim_{x \to1} \dfrac{x + {x}^{2} + {x}^{3} + {x}^{4} - 4}{x -1}$

ANSWER :–

GIVEN :–

• A limit –

$\\ \implies\tt \lim_{x \to1} \dfrac{x + {x}^{2} + {x}^{3} + {x}^{4} - 4}{x -1}$

TO FIND :–

• Value of given limit = ?

SOLUTION :–

• Let the limit –

$\\ \implies\tt P = \lim_{x \to1} \dfrac{x + {x}^{2} + {x}^{3} + {x}^{4} - 4}{x -1}$

• Put the limit –

$\\ \implies\tt P = \dfrac{1+ {(1)}^{2} + {(1)}^{3} + {(1)}^{4} - 4}{1 -1}$

$\\ \implies\tt P = \dfrac{1+1+ 1+1- 4}{1 -1}$

$\\ \implies\tt P = \dfrac{4- 4}{1 -1}$

$\\ \implies\tt P = \dfrac{0}{0}$

• It's an undefined form. Now Let's solve –

$\\ \implies\tt P = \lim_{x \to1} \dfrac{x + {x}^{2} + {x}^{3} + {x}^{4} - 4}{x -1}$

• We can write this as –

$\\ \implies\tt P = \lim_{x \to1} \dfrac{(x - 1)({x}^{3}+2{x}^{2} +3x+4)}{(x -1)}$

$\\ \implies\tt P = \lim_{x \to1} \dfrac{ \cancel{(x - 1)}({x}^{3}+2{x}^{2} +3x+4)}{ \cancel{(x -1)}}$

$\\ \implies\tt P = \lim_{x \to1} ({x}^{3}+2{x}^{2} +3x+4)$

• Now Apply limit –

$\\ \implies\tt P ={(1)}^{3}+2{(1)}^{2} +3(1)+4$

$\\ \implies\tt P =1+2+3+4$

$\\ \implies\tt P =3+7$

$\\ \large \implies{ \boxed{\tt P =10}}$

• Hence , The value of limit is 10.

Answer 2

Answer:

Question :–

• Find the value of given limit –

\begin{gathered} \\ \implies\tt \lim_{x \to1} \dfrac{x + {x}^{2} + {x}^{3} + {x}^{4} - 4}{x -1} \\ \end{gathered}

⟹

x→1

lim

x−1

x+x

2

+x

3

+x

4

−4

ANSWER :–

GIVEN :–

• A limit –

\begin{gathered} \\ \implies\tt \lim_{x \to1} \dfrac{x + {x}^{2} + {x}^{3} + {x}^{4} - 4}{x -1} \\ \end{gathered}

⟹

x→1

lim

x−1

x+x

2

+x

3

+x

4

−4

TO FIND :–

• Value of given limit = ?

SOLUTION :–

• Let the limit –

\begin{gathered} \\ \implies\tt P = \lim_{x \to1} \dfrac{x + {x}^{2} + {x}^{3} + {x}^{4} - 4}{x -1} \\ \end{gathered}

⟹P=

x→1

lim

x−1

x+x

2

+x

3

+x

4

−4

• Put the limit –

\begin{gathered} \\ \implies\tt P = \dfrac{1+ {(1)}^{2} + {(1)}^{3} + {(1)}^{4} - 4}{1 -1} \\ \end{gathered}

⟹P=

1−1

1+(1)

2

+(1)

3

+(1)

4

−4

\begin{gathered} \\ \implies\tt P = \dfrac{1+1+ 1+1- 4}{1 -1} \\ \end{gathered}

⟹P=

1−1

1+1+1+1−4

\begin{gathered} \\ \implies\tt P = \dfrac{4- 4}{1 -1} \\ \end{gathered}

⟹P=

1−1

4−4

\begin{gathered} \\ \implies\tt P = \dfrac{0}{0} \\ \end{gathered}

⟹P=

0

0

• It's an undefined form. Now Let's solve –

\begin{gathered} \\ \implies\tt P = \lim_{x \to1} \dfrac{x + {x}^{2} + {x}^{3} + {x}^{4} - 4}{x -1} \\ \end{gathered}

⟹P=

x→1

lim

x−1

x+x

2

+x

3

+x

4

−4

• We can write this as –

\begin{gathered} \\ \implies\tt P = \lim_{x \to1} \dfrac{(x - 1)({x}^{3}+2{x}^{2} +3x+4)}{(x -1)} \\ \end{gathered}

⟹P=

x→1

lim

(x−1)

(x−1)(x

3

+2x

2

+3x+4)

\begin{gathered} \\ \implies\tt P = \lim_{x \to1} \dfrac{ \cancel{(x - 1)}({x}^{3}+2{x}^{2} +3x+4)}{ \cancel{(x -1)}} \\ \end{gathered}

⟹P=

x→1

lim

(x−1)

(x−1)

(x

3

+2x

2

+3x+4)

\begin{gathered} \\ \implies\tt P = \lim_{x \to1} ({x}^{3}+2{x}^{2} +3x+4) \\ \end{gathered}

⟹P=

x→1

lim

(x

3

+2x

2

+3x+4)

• Now Apply limit –

\begin{gathered} \\ \implies\tt P ={(1)}^{3}+2{(1)}^{2} +3(1)+4\\ \end{gathered}

⟹P=(1)

3

+2(1)

2

+3(1)+4

\begin{gathered} \\ \implies\tt P =1+2+3+4\\ \end{gathered}

⟹P=1+2+3+4

\begin{gathered} \\ \implies\tt P =3+7\\ \end{gathered}

⟹P=3+7

\begin{gathered} \\ \large \implies{ \boxed{\tt P =10}}\\ \end{gathered}

⟹

P=10

• Hence , The value of limit is 10.