Question

Evaluate:


`\\ { \displaystyle{ \rm \lim_(x \to1) \left \{ \frac{ {x}^(4 ) - {3x}^(2) + 2 }{ {x}^(3) - {5x}^(2) + 3x + 1} \right \} }}`

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Answer 1

Without using L'Hopital's rule:

Factorize the numerator.

`x^4 - 3x^2 + 2 = (x^2 - 2) (x^2 - 1) = (x^2 - 2) (x + 1) (x - 1)`

Factorizing cubics is usually a bit trickier. But we know x = 1 makes the denominator vanish, so x - 1 is a factor of the cubic. So for some constants a and b,

`x^3 - 5x^2 + 3x + 1 = (x - 1)^3 + a(x - 1)^2 + b(x - 1)`

Expanding the right side gives

`x^3 - 5x^2 + 3x + 1 = x^3 + (a - 3)x^2 + (-2a + b + 3)x + a - b - 1`

Then

`a - 3 = -5 \implies a = -2`

`-2a + b + 3 = 3 \implies b = -4`

So we rewrite the limit as

`\displaystyle \lim_(x\to1) (x^4 - 3x^2 + 2)/(x^3 - 5x^2 + 3x + 1) = \lim_(x\to1) ((x^2 - 2) (x + 1) (x - 1))/((x-1)^3 - 2(x-1)^2 - 4(x-1))`

x is approaching 1 so x ≠ 1 and we can cancel out factors of x - 1 :

`\displaystyle \lim_(x\to1) (x^4 - 3x^2 + 2)/(x^3 - 5x^2 + 3x + 1) = \lim_(x\to1) ((x^2 - 2) (x + 1))/((x-1)^2 - 2(x-1) - 4)`

The simplified limand is continuous at x = 1, so we can now evaluate the limit by direct substitution.

`\displaystyle \lim_(x\to1) (x^4 - 3x^2 + 2)/(x^3 - 5x^2 + 3x + 1) = ((1^2 - 2) (1 + 1))/((1-1)^2 - 2(1-1) - 4) = -\frac24 = \boxed{-\frac12}`

Answer 2

To Evaluate:

`\\ { \displaystyle{ \rm \lim_(x \to1) \left \{ \frac{ {x}^(4 ) - {3x}^(2) + 2 }{ {x}^(3) - {5x}^(2) + 3x + 1} \right \} }}\\ \\`

Using L'Hospital's Rule

Let f(x) and g(x) be two common functions which are differentiable on an open Interval , at a point a where,

`\\{ \displaystyle{ \lim_(x \to a){ \rm{f(x)} = { \displaystyle{ \lim_(x \to a)}} \: { \rm{g(x)}} = 0 \: or \: \pm \infty}}}\\ \\`

Then,

`\\{ \displaystyle{ \lim_(x \to a) \frac{{ \rm{f (x)}}}{{ \rm{g(x)}}} = { \displaystyle{ \lim_(x \to a )}} \frac{{ \rm{f '(x)}}}{{ \rm{g'(x)}}}}}\\ \\`

As x to 0 , we have:

`\\{ \displaystyle{ \lim_(x \to 1)}} \left( \large\rm\frac{ {x}^(4) - {3x}^(2) + 2 }{ {x}^(3) - {5x}^(2) + 3x + 1} \right) = (0)/(0)\\ \\`

Therefore,

`\\{ \large{ \displaystyle{ \lim_(x \to1)}} \left( \large \rm\frac{ {x}^(4) - {3x}^(2) + 2}{ {x}^(3) - {5x}^(2) + 3x + 1 } \right) = { \displaystyle{ \lim_(x \to1) { \small{\frac{ {4x}^(3) - 6x}{ {3x}^(2) - 10x + 3 }}} = { \small{(4 - 6)/(3 - 10 + 3) }} = - (2)/( - 4) = - (1)/(2) }}}\\ \\`

Hence,

`\\{ \displaystyle{ \lim_(x \to1){\rm{ \left( \frac{ {x}^(4) - 3 {x}^(2) + 2 }{ {x}^(3) - 5 {x}^(2) + 3x + 1 } \right) = { \boxed{ \red{ - (1)/(2) }}}}}}}\\ \\`