Find the limit algebraically.
-$$\lim _ { x \rightarrow 1 } \frac { x ^ { 3 } + 5 x ^ { 2 } + 3 x - 9 } { x - 1 }$$
A) -16
B) 0
C) 16
D) does not exist

Question

Quizplus · Accepted Answer

To solve this limit algebraically, we can factor the numerator and cancel out a common factor with the denominator. The given expression is:$$\lim _ { x \rightarrow 1 } \frac { x ^ { 3 } + 5 x ^ { 2 } + 3 x - 9 } { x - 1 }$$First, notice that direct substitution of $x = 1$ into the expression results in a $0/0$ form, indicating we should factor the numerator if possible. To factor the numerator, we look for a polynomial that has $x - 1$ as a factor, since that's our denominator. The numerator can be factored by polynomial division or by using synthetic division, aiming to divide by $x - 1$. Doing so, we find:$$x ^ { 3 } + 5 x ^ { 2 } + 3 x - 9 = (x - 1)(x^2 + 6x + 9)$$Notice that $x^2 + 6x + 9$ can be further factored into $(x + 3)^2$, but that's not necessary for solving this limit. Now, the expression becomes:$$\lim _ { x \rightarrow 1 } \frac { (x - 1)(x^2 + 6x + 9) } { x - 1 }$$Canceling $x - 1$ from the numerator and the denominator, we get:$$\lim _ { x \rightarrow 1 } (x^2 + 6x + 9)$$Now, we can directly substitute $x = 1$ into the simplified expression:$$1^2 + 6(1) + 9 = 1 + 6 + 9 = 16$$Therefore, the limit is 16.

Find the Limit Algebraically lim⁡x→1x3+5x2+3x−9x−1\lim _ { x \rightarrow 1 } \frac { x ^ { 3 } + 5 x ^ { 2 } + 3 x - 9 } { x - 1 }x→1lim​x−1x3+5x2+3x−9​

Find the Limit Algebraically $\lim _ { x \rightarrow 1 } \frac { x ^ { 3 } + 5 x ^ { 2 } + 3 x - 9 } { x - 1 }$