Answer:
Given that 1 and 4 are vertical asymtotes we have;
(a) -∞
(b) +∞
(c) +∞
(d) -∞
Step-by-step explanation:
(a) For the function;
[tex]\lim_{x\rightarrow 4^{-}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{x^{2}-5\cdot x+4} \right )[/tex]
We have the denominator given by the expression, x² - 5·x + 4 which can be factorized as (x - 4)(x - 1)
Therefore, as the function approaches 4 from the left [lim (x → 4⁻)] gives;
[tex]\lim_{x\rightarrow 4^{-}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{(x - 1)\cdot (x - 4)} \right )[/tex] [tex]\lim_{x\rightarrow 4^{-}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{(3.999 - 1)\cdot (3.999 - 4)} \right )[/tex] [tex]\lim_{x\rightarrow 4^{-}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{(2.999)\cdot (-0.001)} \right )[/tex][tex]=- \infty[/tex]
(b) Similarly, we have;
[tex]\lim_{x\rightarrow 4^{+}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{x^{2}-5\cdot x+4} \right )[/tex]
We have the denominator given by the expression, x² - 5·x + 4 which can be factorized as (x - 4)(x - 1)
Therefore, as the function approaches 4 from the right [lim (x → 4⁺)] gives;
[tex]\lim_{x\rightarrow 4^{+}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{(x - 1)\cdot (x - 4)} \right )[/tex] [tex]\lim_{x\rightarrow 4^{+}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{(4.0001 - 1)\cdot (4.0001 - 4)} \right )[/tex] [tex]\lim_{x\rightarrow 4^{+}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{(3.0001)\cdot (0.0001)} \right )[/tex][tex]= +\infty[/tex]
(c)
[tex]\lim_{x\rightarrow 1^{-}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{x^{2}-5\cdot x+4} \right )[/tex]
We have the denominator given by the expression, x² - 5·x + 4 which can be factorized as (x - 4)(x - 1)
Therefore, as the function approaches 1 from the left [lim (x → 1⁻)] gives;
[tex]\lim_{x\rightarrow 1 ^{-}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{(x - 1)\cdot (x - 4)} \right )[/tex] [tex]\lim_{x\rightarrow 1^{-}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{(0.999 - 1)\cdot (0.999 - 4)} \right )[/tex] [tex]\lim_{x\rightarrow 4^{+}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{(-0.001)\cdot (-3.001)} \right )[/tex][tex]=+ \infty[/tex]
(d) As the function approaches 1 from the right [lim (x → 1⁺)]
We have;
[tex]\lim_{x\rightarrow 1^{+}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{(1.0001 - 1)\cdot (1.0001 - 4)} \right )[/tex]= [tex]\lim_{x\rightarrow 1^{+}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{(0.0001)\cdot (-2.999)} \right ) =- \infty[/tex]
