Math Problem Statement

Solve the limit problems provided in the image.

Solution

You have provided an image with several limit problems. I will solve these one by one, providing step-by-step explanations for each. Let's begin with the first few.


1) limx1x2+3x+15x2x+1\lim_{x \to 1} \frac{\sqrt{x^2 + 3x + 1} - 5x}{2x + 1}

To solve this, let's first try direct substitution by plugging in x=1x = 1:

Numerator:12+3(1)+15(1)=1+3+15=55\text{Numerator:} \quad \sqrt{1^2 + 3(1) + 1} - 5(1) = \sqrt{1 + 3 + 1} - 5 = \sqrt{5} - 5 Denominator:2(1)+1=3\text{Denominator:} \quad 2(1) + 1 = 3

Thus, the limit becomes: limx1553\lim_{x \to 1} \frac{\sqrt{5} - 5}{3}

Since this is a constant, the limit is simply:

553\boxed{\frac{\sqrt{5} - 5}{3}}


2) limx3x3+5x+12x3+x29\lim_{x \to \infty} \frac{3x^3 + 5x + 1}{2x^3 + x^2 - 9}

We will divide the numerator and the denominator by x3x^3 (the highest power of xx) to simplify:

limx3x3x3+5xx3+1x32x3x3+x2x39x3=limx3+5x2+1x32+1x9x3\lim_{x \to \infty} \frac{\frac{3x^3}{x^3} + \frac{5x}{x^3} + \frac{1}{x^3}}{\frac{2x^3}{x^3} + \frac{x^2}{x^3} - \frac{9}{x^3}} = \lim_{x \to \infty} \frac{3 + \frac{5}{x^2} + \frac{1}{x^3}}{2 + \frac{1}{x} - \frac{9}{x^3}}

As xx \to \infty, all terms with 1xn\frac{1}{x^n} vanish, leaving:

limx32=32\lim_{x \to \infty} \frac{3}{2} = \boxed{\frac{3}{2}}


3) limx2x3+8x23x10\lim_{x \to 2} \frac{x^3 + 8}{x^2 - 3x - 10}

First, factor the denominator:

x23x10=(x5)(x+2)x^2 - 3x - 10 = (x - 5)(x + 2)

Substitute x=2x = 2 directly:

Numerator:23+8=8+8=16\text{Numerator:} \quad 2^3 + 8 = 8 + 8 = 16 Denominator:(25)(2+2)=(3)(4)=12\text{Denominator:} \quad (2 - 5)(2 + 2) = (-3)(4) = -12

Thus, the limit is:

limx21612=43\lim_{x \to 2} \frac{16}{-12} = \boxed{-\frac{4}{3}}


4) limx1x1x1\lim_{x \to 1} \frac{\sqrt{x} - 1}{x - 1}

This is a classic indeterminate form. We can use the conjugate to simplify:

Multiply numerator and denominator by the conjugate of the numerator:

x1x1×x+1x+1=(x)212(x1)(x+1)=x1(x1)(x+1)\frac{\sqrt{x} - 1}{x - 1} \times \frac{\sqrt{x} + 1}{\sqrt{x} + 1} = \frac{(\sqrt{x})^2 - 1^2}{(x - 1)(\sqrt{x} + 1)} = \frac{x - 1}{(x - 1)(\sqrt{x} + 1)}

Now cancel x1x - 1:

1x+1\frac{1}{\sqrt{x} + 1}

Substitute x=1x = 1:

11+1=12\frac{1}{\sqrt{1} + 1} = \frac{1}{2}

Thus, the limit is:

12\boxed{\frac{1}{2}}


Would you like me to continue solving the rest of the problems in the image? Please let me know!

Here are 5 related questions you might find helpful:

  1. How do you handle indeterminate forms like 0/00/0?
  2. What are common techniques to simplify complex limit problems?
  3. How can the conjugate method help in solving limits involving square roots?
  4. What is L'Hopital's rule and when should it be used?
  5. How do we approach limits involving infinity in rational functions?

Tip: Always check for factorizations and simplifications before applying advanced techniques like L'Hopital's rule.

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Math Problem Analysis

Mathematical Concepts

Limits
Algebra
Rational Functions
Indeterminate Forms
L'Hopital's Rule
Conjugates

Formulas

Limit definition: lim(x → c) f(x)
L'Hopital's rule: lim(x → c) f(x)/g(x) = lim(x → c) f'(x)/g'(x) if it leads to an indeterminate form
Conjugate method: Multiply by (sqrt(a) + sqrt(b)) to simplify square root expressions

Theorems

Squeeze Theorem
L'Hopital's Rule
Limit properties (e.g., sum, product, and quotient rules)

Suitable Grade Level

Undergraduate Calculus