Question

Evaluate


`lim_(x→1)( \frac{ {ax}^(2) + bx + c }{c {x}^(2) + bx + a } )`
Please help​

Answer 1

`{ \red{ { \sf{1}}}}`

Step-by-step explanation:

`{ \green{ \sf{ lim_(x→1)( \frac{a {x}^(2) + bx + c }{c {x}^(2) + bx + a } ) }}}`

`{ \green{ \sf{ \frac{a {(1)}^(2) + b(1) + c}{c ({1})^(2) + b(1) + a}}}}`

`{ \green{ \sf{(a + b + c)/(c + b + a)}}}`

`= 1`

Answer 2

Answer: 1

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Work Shown:

`\displaystyle L = \lim_(x \to 1) \left(\frac{a\text{x}^2+b\text{x}+c}{c\text{x}^2+b\text{x}+a}\right)\\\\\\\displaystyle L = (a(1)^2+b(1)+c)/(c(1)^2+b(1)+a)\\\\\\\displaystyle L = (a+b+c)/(c+b+a)\\\\\\\displaystyle L = (a+b+c)/(a+b+c)\\\\\\\displaystyle L = 1\\\\\\`

Therefore,

`\displaystyle \lim_(x \to 1) \left(\frac{a\text{x}^2+b\text{x}+c}{c\text{x}^2+b\text{x}+a}\right)=1`

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Step-by-step explanation:

Apply substitution. Replace each copy of x with 1. We don't have to worry about division by zero errors in this case. After replacing x, we have a+b+c in the numerator and denominator, in which cancel to get us 1 as the final answer.