1) $$ \lim _{x \rightarrow-2}\left(3 x^{3}+5 x^{2}-1\right) $$ 2) $$ \lim _{x \rightarrow 0}\left(\frac{x e^{-x}}{3 e^{x}-3}\right) $$ 3) $$ \lim _{x \rightarrow-5} \frac{x^{2}-25}{(x+5)(x-6)} $$

Question

UpStudy AI · Accepted Answer

Let's evaluate each of the given limits step by step.

---

### **1) $$ \lim _{x \rightarrow-2}\left(3 x^{3}+5 x^{2}-1\right) $$**

**Solution:**

Since the function $$ f(x) = 3x^3 + 5x^2 - 1 $$ is a polynomial, it's continuous everywhere. Therefore, the limit as $$ x $$ approaches $$-2$$ is simply the value of the function at $$ x = -2 $$.

$$
\begin{align*}
f(-2) &= 3(-2)^3 + 5(-2)^2 - 1 \
&= 3(-8) + 5(4) - 1 \
&= -24 + 20 - 1 \
&= -5
\end{align*}
$$

**Answer:** $$-5$$

---

### **2) $$ \lim _{x \rightarrow 0}\left(\frac{x e^{-x}}{3 e^{x}-3}\right) $$**

**Solution:**

First, evaluate the limit by direct substitution to check the form:

$$
\begin{align*}
\text{Numerator as } x \to 0: &\quad x e^{-x} \to 0 \times 1 = 0 \
\text{Denominator as } x \to 0: &\quad 3 e^{x} - 3 \to 3(1) - 3 = 0 \
\end{align*}
$$

This results in an indeterminate form $$ \frac{0}{0} $$. To resolve this, apply **L'Hospital's Rule**, which states that if $$ \lim_{x \to c} \frac{f(x)}{g(x)} $$ is of the form $$ \frac{0}{0} $$ or $$ \frac{\infty}{\infty} $$, then:

$$
\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}
$$

**Compute the derivatives:**

- **Numerator:** $$ f(x) = x e^{-x} $$
  
  $$
  f'(x) = e^{-x} + x (-e^{-x}) = e^{-x} - x e^{-x}
  $$

- **Denominator:** $$ g(x) = 3 e^{x} - 3 $$
  
  $$
  g'(x) = 3 e^{x}
  $$

**Apply L'Hospital's Rule:**

$$
\lim_{x \to 0} \frac{e^{-x} - x e^{-x}}{3 e^{x}} = \frac{e^{0} - 0 \times e^{0}}{3 e^{0}} = \frac{1}{3}
$$

**Answer:** $$\frac{1}{3}$$

---

### **3) $$ \lim _{x \rightarrow-5} \frac{x^{2}-25}{(x+5)(x-6)} $$**

**Solution:**

First, factor the numerator:

$$
x^2 - 25 = (x - 5)(x + 5)
$$

So the expression becomes:

$$
\frac{(x - 5)(x + 5)}{(x + 5)(x - 6)} = \frac{(x - 5)}{(x - 6)} \quad \text{for} \quad x \neq -5
$$

Now, cancel the common factor $$ (x + 5) $$:

$$
\lim_{x \to -5} \frac{x - 5}{x - 6}
$$

Substitute $$ x = -5 $$:

$$
\frac{-5 - 5}{-5 - 6} = \frac{-10}{-11} = \frac{10}{11}
$$

**Answer:** $$\frac{10}{11}$$

---

**Summary of Answers:**

1. $$-5$$
2. $$\dfrac{1}{3}$$
3. $$\dfrac{10}{11}$$
The answers are: 1. $$-5$$ 2. $$\frac{1}{3}$$ 3. $$\frac{10}{11}$$

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Calculus Questions