Math Problem Statement

in no. 2 the exponent of e is t

In No. 2, the exponent of e is t.

Solution

Thank you for the clarification! Let's analyze and solve the problems from the chalkboard.


1. limx0sin2xx\lim_{x \to 0} \frac{\sin^2{x}}{x}

To solve: limx0sin2xx=limx0sinxsinxx.\lim_{x \to 0} \frac{\sin^2{x}}{x} = \lim_{x \to 0} \frac{\sin{x} \cdot \sin{x}}{x}. Using the standard limit property limx0sinxx=1\lim_{x \to 0} \frac{\sin{x}}{x} = 1: limx0sin2xx=limx0sinxlimx0sinxx=10=0.\lim_{x \to 0} \frac{\sin^2{x}}{x} = \lim_{x \to 0} \sin{x} \cdot \lim_{x \to 0} \frac{\sin{x}}{x} = 1 \cdot 0 = 0. Final Answer: 0


2. limx0et1t+x+13x+14x+3\lim_{x \to 0} \frac{e^t - 1}{t} + \frac{x + 13x + 14}{x + 3}

Simplify:

Let's go step by step with each part of the problem: --- ### 1. $$\lim_{x \to 0} \frac{\sin^2{x}}{x}$$ This can be written as: \[ \lim_{x \to 0} \frac{\sin^2{x}}{x} = \lim_{x \to 0} \sin{x} \cdot \lim_{x \to 0} \frac{\sin{x}}{x}.$$ Using the standard limit property $$\lim_{x \to 0} \frac{\sin{x}}{x} = 1$$: $$\lim_{x \to 0} \frac{\sin^2{x}}{x} = \lim_{x \to 0} \sin{x} \cdot 1 = 0 \cdot 1 = 0.$$ **Answer**: $$0$$ --- ### 2. $$\lim_{x \to 0} \frac{e^t - 1}{t} + \frac{x + 13x + 14}{x + 3}$$ #### Part 1: $$\lim_{t \to 0} \frac{e^t - 1}{t}$$ Using the standard limit property for exponential functions: $$\lim_{t \to 0} \frac{e^t - 1}{t} = 1.$$ #### Part 2: $$\frac{x + 13x + 14}{x + 3}$$ Simplify the numerator: $$x + 13x + 14 = 14x + 14 = 14(x + 1).$$ So the expression becomes: $$\frac{14(x + 1)}{x + 3}.$$ Since there is no indeterminate form as $$x \to 0$$, simply substitute $$x = 0$$: $$\frac{14(0 + 1)}{0 + 3} = \frac{14}{3}.$$ Adding the two results: $$\lim_{x \to 0} \frac{e^t - 1}{t} + \frac{x + 13x + 14}{x + 3} = 1 + \frac{14}{3} = \frac{17}{3}.$$ **Answer**: $$\frac{17}{3}$$ --- ### 3. $$\lim_{x \to 3} \frac{x^2 - 9}{\sqrt{x} + 3}$$ Factorize the numerator: $$x^2 - 9 = (x - 3)(x + 3).$$ The expression becomes: $$\frac{(x - 3)(x + 3)}{\sqrt{x} + 3}.$$ Cancel out $$x - 3$$ since $$x \to 3$$ and not $$x = 3$$: $$\lim_{x \to 3} \frac{(x - 3)(x + 3)}{\sqrt{x} + 3} = \lim_{x \to 3} \frac{x + 3}{\sqrt{x} + 3}.$$ Substitute $$x = 3$$: $$\frac{3 + 3}{\sqrt{3} + 3} = \frac{6}{\sqrt{3} + 3}.$$ Rationalize the denominator: $$\frac{6}{\sqrt{3} + 3} \cdot \frac{\sqrt{3} - 3}{\sqrt{3} - 3} = \frac{6(\sqrt{3} - 3)}{(\sqrt{3} + 3)(\sqrt{3} - 3)} = \frac{6(\sqrt{3} - 3)}{3 - 9} = \frac{6(\sqrt{3} - 3)}{-6}.$$ Simplify: $$\frac{6(\sqrt{3} - 3)}{-6} = -(\sqrt{3} - 3) = -\sqrt{3} + 3.$$ **Answer**: $$-\sqrt{3} + 3$$ --- ### Summary of Results 1. $$\lim_{x \to 0} \frac{\sin^2{x}}{x} = 0$$ 2. $$\lim_{x \to 0} \frac{e^t - 1}{t} + \frac{x + 13x + 14}{x + 3} = \frac{17}{3}$$ 3. $$\lim_{x \to 3} \frac{x^2 - 9}{\sqrt{x} + 3} = -\sqrt{3} + 3$$ --- Would you like further explanation or step-by-step details for any part? ### Related Questions 1. What are the standard limits for trigonometric and exponential functions? 2. How do you rationalize the denominator of a fraction? 3. What is L'Hôpital's Rule, and when is it applicable for limits? 4. Can you explain the concept of factoring polynomials in limit problems? 5. How is substitution used in solving limits? **Tip**: When solving limits involving indeterminate forms, always check for factorization, rationalization, or standard limit properties first before applying advanced methods.

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Math Problem Analysis

Mathematical Concepts

Limits
Exponential Functions
Simplification
Rationalization
Polynomial Factorization

Formulas

lim_{x \to 0} \frac{sin(x)}{x} = 1
lim_{t \to 0} \frac{e^t - 1}{t} = 1
Polynomial factorization: x^2 - 9 = (x - 3)(x + 3)

Theorems

Limit properties of trigonometric and exponential functions
Rationalization of square root denominators
Polynomial Factorization Theorem

Suitable Grade Level

Grades 11-12 (Precalculus and Calculus)