Evaluate:
[tex] \\ { \displaystyle{ \rm \lim_{x \to1} \left \{ \frac{ {x}^{4 } - {3x}^{2} + 2 }{ {x}^{3} - {5x}^{2} + 3x + 1} \right \} }} [/tex]
[tex] \: [/tex]
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Answer:

To Evaluate:

[tex] \\ { \displaystyle{ \rm \lim_{x \to1} \left \{ \frac{ {x}^{4 } - {3x}^{2} + 2 }{ {x}^{3} - {5x}^{2} + 3x + 1} \right \} }}\\ \\[/tex]

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,

[tex]\\{ \displaystyle{ \lim_{x \to a}{ \rm{f(x)} = { \displaystyle{ \lim_{x \to a}}} \: { \rm{g(x)}} = 0 \: or \: \pm \infty}}}\\ \\[/tex]

Then,

[tex]\\{ \displaystyle{ \lim_{x \to a} \frac{{ \rm{f (x)}}}{{ \rm{g(x)}}} = { \displaystyle{ \lim_{x \to a }}} \frac{{ \rm{f '(x)}}}{{ \rm{g'(x)}}}}}\\ \\[/tex]

As x to 0 , we have:

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

Therefore,

[tex]\\{ \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{\frac{4 - 6}{3 - 10 + 3} }} = - \frac{2}{ - 4} = - \frac{1}{2} }}}\\ \\[/tex]

Hence,

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

Without using L'Hopital's rule:

Factorize the numerator.

[tex]x^4 - 3x^2 + 2 = (x^2 - 2) (x^2 - 1) = (x^2 - 2) (x + 1) (x - 1)[/tex]

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,

[tex]x^3 - 5x^2 + 3x + 1 = (x - 1)^3 + a(x - 1)^2 + b(x - 1)[/tex]

Expanding the right side gives

[tex]x^3 - 5x^2 + 3x + 1 = x^3 + (a - 3)x^2 + (-2a + b + 3)x + a - b - 1[/tex]

Then

[tex]a - 3 = -5 \implies a = -2[/tex]

[tex]-2a + b + 3 = 3 \implies b = -4[/tex]

So we rewrite the limit as

[tex]\displaystyle \lim_{x\to1} \frac{x^4 - 3x^2 + 2}{x^3 - 5x^2 + 3x + 1} = \lim_{x\to1} \frac{(x^2 - 2) (x + 1) (x - 1)}{(x-1)^3 - 2(x-1)^2 - 4(x-1)}[/tex]

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

[tex]\displaystyle \lim_{x\to1} \frac{x^4 - 3x^2 + 2}{x^3 - 5x^2 + 3x + 1} = \lim_{x\to1} \frac{(x^2 - 2) (x + 1)}{(x-1)^2 - 2(x-1) - 4}[/tex]

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

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

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