Question

Compute the following limits.
2x +1
lim
x-1( 3x2 - 4x +5​

Answer 1

Correct question: Compute the limit of the following.

$\lim_{x \to 1} \frac{2x+1}{3x^2-4x+5}$

Answer:

The value of the limit $\lim_{x \to 1} \frac{2x+1}{3x^2-4x+5}$ is $\frac{3}{4}$.

Step-by-step explanation:

Consider the limit as follows:

$\lim_{x \to 1} \frac{2x+1}{3x^2-4x+5}$ ...... (1)

Here, $f(x)= \frac{2x+1}{3x^2-4x+5}$

We have to find the limit when $x \to 1$, i.e., $x$ approaches to $1$.

For this, substitute the value $1$ for $x$ in the equation (1) as follows:

$\lim_{x \to 1} \frac{2x+1}{3x^2-4x+5}=\frac{2(1)+1}{3(1)^2-4(1)+5}$

Simplify as follows:

$\lim_{x \to 1} \frac{2x+1}{3x^2-4x+5}=\frac{2+1}{3-4+5}$

                          $=\frac{3}{4}$ (finite number)

Therefore, the limit exists and the value of $\lim_{x \to 1} \frac{2x+1}{3x^2-4x+5}$ is $\frac{3}{4}$.

#SPJ6

Answer 2

Answer:

3/4

Step-by-step explanation:

directly substitute the x value we get 2(1)+1/3-4+5=3/4