Question

Use the limit theorem and the properties of limits to find the limit.
Picture provided below

Answer 1

Answer:

b. 1/2

Step-by-step explanation:

lim        (x -3)(x +2)

x-->-∞    ---------------

              2x^2 + x +1

= lim        (x^2 -3x +2x - 6)

x-->-∞    -----------------------

              2x^2 + x +1

= lim        (x^2 -x - 6)

x-->-∞    -----------------------

              2x^2 + x +1

When we plug in x = -∞, we get indeterminate form.

Now we have to use the L'hospital rule.

d/dx (x^2 - x - 6) = 2x -1

d/dx (2x^2 + x + 1) = 4x + 1

Now apply the limit

lim            (2x - 1) / (4x + 1)

x--->-∞

Here we have to degree of the numerator and the denominator of the same. In this case, if x --> -∞, we get the result as the coefficient of the leading term as the result.

According to the rule, we get

= 2/4

Which can simplified as 1/2

The answer is 1/2

Hope this will helpful.

Thank you.

Answer 2

Answer:

Option B

Step-by-step explanation:

We know that the limit when x tends to infinity of:

$\frac{1}{x ^ n}$ is approximately zero when n is a positive real number. The term of greatest exponent in the function is $x ^ 2$

Based on this we do the following.

Divide each term of the numerator and denominator between the term with the greatest exponent of the expression. Then it is:

$\lim_{x\to \infty}\frac{(x-3)(x+2)}{2x^2 + x +1}\\\\= \lim_{x\to \infty} \frac{(x^2 -x -6)}{2x^2 + x +1}\\\\\\ \lim_{x \to \infty}\frac{\frac{x^2}{x^2} -\frac{x}{x^2} -\frac{6}{x^2}}{2\frac{x^2}{x^2} + \frac{x}{x^2} +\frac{1}{x^2}}$

Then as the $\frac{1}{\infty} \to 0$  then we have left:

$= \lim_{x \to \infty} \frac{\frac{x^2}{x^2}}{2\frac{x^2}{x^2}}\\\\\lim_{x \to \infty}\frac{1}{2} = \frac{1}{2}$

The answer is: Option b