Question

Evaluate each limit. Give exact answers.

Answer 1

Answer:

Given that 1 and 4 are vertical asymtotes we have;

(a) -∞

(b) +∞

(c) +∞

(d) -∞

Step-by-step explanation:

(a) For the function;

$\lim_{x\rightarrow 4^{-}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{x^{2}-5\cdot x+4} \right )$

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;

$\lim_{x\rightarrow 4^{-}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{(x - 1)\cdot (x - 4)} \right )$ $\lim_{x\rightarrow 4^{-}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{(3.999 - 1)\cdot (3.999 - 4)} \right )$ $\lim_{x\rightarrow 4^{-}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{(2.999)\cdot (-0.001)} \right )$$=- \infty$

(b) Similarly, we have;

$\lim_{x\rightarrow 4^{+}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{x^{2}-5\cdot x+4} \right )$

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;

$\lim_{x\rightarrow 4^{+}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{(x - 1)\cdot (x - 4)} \right )$ $\lim_{x\rightarrow 4^{+}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{(4.0001 - 1)\cdot (4.0001 - 4)} \right )$ $\lim_{x\rightarrow 4^{+}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{(3.0001)\cdot (0.0001)} \right )$$= +\infty$

(c)

$\lim_{x\rightarrow 1^{-}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{x^{2}-5\cdot x+4} \right )$

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;

$\lim_{x\rightarrow 1 ^{-}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{(x - 1)\cdot (x - 4)} \right )$ $\lim_{x\rightarrow 1^{-}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{(0.999 - 1)\cdot (0.999 - 4)} \right )$ $\lim_{x\rightarrow 4^{+}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{(-0.001)\cdot (-3.001)} \right )$$=+ \infty$

(d) As the function approaches 1 from the right [lim (x → 1⁺)]

We have;

$\lim_{x\rightarrow 1^{+}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{(1.0001 - 1)\cdot (1.0001 - 4)} \right )$= $\lim_{x\rightarrow 1^{+}}\left (\dfrac{2\cdot x^{2}+13\cdot x+20}{(0.0001)\cdot (-2.999)} \right ) =- \infty$