Question

please answer this question​

Answer 1

Answer:

$\displaystyle \lim_{x \to 2} \frac{\sqrt{x + 7} - 3 \sqrt{2x - 3}}{\sqrt[3]{x + 6} - 2 \sqrt[3]{3x - 5}} = \boxed{ \frac{34}{23} }$

General Formulas and Concepts:
Calculus

Limits

Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to c} x = c$

Limit Property [Multiplied Constant]:
$\displaystyle \lim_{x \to c} bf(x) = b \lim_{x \to c} f(x)$

Limit Property [Multiplication]:
$\displaystyle \lim_{x \to c} f(x)g(x) = \lim_{x \to c} f(x) \lim_{x \to c} g(x)$

Step-by-step explanation:

*Note:

The problem is too big to fit all work. I will assume that you know how to do basic algebra and calculus.

Step 1: Define

Identify given limit.

$\displaystyle \lim_{x \to 2} \frac{\sqrt{x + 7} - 3 \sqrt{2x - 3}}{\sqrt[3]{x + 6} - 2 \sqrt[3]{3x - 5}}$

Step 2: Find Limit (Algebriaclly)

Rationalize the function and apply basic limit techniques listed under "Calculus":

$\displaystyle\begin{aligned}\lim_{x \to 2} \frac{\sqrt{x + 7} - 3 \sqrt{2x - 3}}{\sqrt[3]{x + 6} - 2 \sqrt[3]{3x - 5}} & = \lim_{x \to 2} - \frac{17(x - 2)}{\big( \sqrt[3]{x + 6} - 2 \sqrt[3]{3x - 5} \big) \big( \sqrt{x + 7} + 3 \sqrt{2x - 3} \big)} \\& = -17 \lim_{x \to 2} \frac{1}{\sqrt{x + 7} + 3 \sqrt{2x - 3}} \lim_{x \to 2} \frac{x - 2}{\sqrt[3]{x + 6} - 2 \sqrt[3]{3x - 5}}\end{aligned}$

$\displaystyle\begin{aligned}\lim_{x \to 2} \frac{\sqrt{x + 7} - 3 \sqrt{2x - 3}}{\sqrt[3]{x + 6} - 2 \sqrt[3]{3x - 5}} & = - \frac{17}{6} \lim_{x \to 2} \frac{x - 2}{\sqrt[3]{x + 6} - 2 \sqrt[3]{3x - 5}} \\\end{aligned}$

Let's now focus on just the rationalization of the function within the limit:

$\displaystyle\begin{aligned}\frac{x - 2}{\sqrt[3]{x + 6} - 2 \sqrt[3]{3x - 5}} & = \frac{(x - 2) \bigg[ (x + 6)^\Big{\frac{2}{3}} + 2 \sqrt[3]{x + 6} \sqrt[3]{3x - 5} + 4(3x - 5)^\Big{\frac{2}{3}} \bigg] }{\Big( \sqrt[3]{x + 6} - 2 \sqrt[3]{3x - 5} \Big) \bigg[ (x + 6)^\Big{\frac{2}{3}} + 2 \sqrt[3]{x + 6} \sqrt[3]{3x - 5} + 4(3x - 5)^\Big{\frac{2}{3}} \bigg] } \\& = - \frac{1}{23} \bigg[ (x + 6)^\Big{\frac{2}{3}} + 2 \sqrt[3]{x + 6} \sqrt[3]{3x - 5} + 4(3x - 5)^\Big{\frac{2}{3}} \bigg] \\\end{aligned}$

Substitute in the simplified function and continue:

$\displaystyle\begin{aligned}\lim_{x \to 2} \frac{\sqrt{x + 7} - 3 \sqrt{2x - 3}}{\sqrt[3]{x + 6} - 2 \sqrt[3]{3x - 5}} & = - \frac{17}{6} \lim_{x \to 2} \frac{x - 2}{\sqrt[3]{x + 6} - 2 \sqrt[3]{3x - 5}} \\& = \frac{17}{138} \lim_{x \to 2} \bigg[ (x + 6)^\Big{\frac{2}{3}} + 2 \sqrt[3]{x + 6} \sqrt[3]{3x - 5} + 4(3x - 5)^\Big{\frac{2}{3}} \bigg] \\& = \frac{17}{138} (12) \\& = \boxed{ \frac{34}{23} } \\\end{aligned}$

∴ we have evaluated the following limit without L' Hopital's Rule.

Step 3: Geometrical Interpretation

We can also graph the given function and approximate the limit. Please refer to the attachment for visualization.

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Learn more about limits: https://brainly.com/question/27593180

Learn more about Calculus: https://brainly.com/question/27805589

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Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

Answer 2

Answer:

  34/23

Step-by-step explanation:

The limit can be evaluated by simplifying the expression in such a way as to cancel the offending zeros.

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Define the following (to help keep the algebra cleaner):

  $a=\sqrt{x+7},\ b=3\sqrt{2x-3}\\\\c=\sqrt[3]{x+6}.\ d=2\sqrt[3]{3x-5}$

Then the desired limit is ...

  $\displaystyle \lim_{x\to 2}\dfrac{a-b}{c-d}$

We observe that ...

  $(a-b)(a+b)=\underline{a^2-b^2}=(x+7)-(3^2(2x-3))=-17x+34=\underline{-17(x-2)}\\\\(c-d)(c^2+cd+d^2)=\underline{c^3-d^3}=(x+6)-(2^3(3x-5))=-23x+46=\underline{-23(x-2)}\\\\\displaystyle\lim_{x\to2}{(a+b)}=3+3=6\\\\\lim_{x\to2}{(c^2+cd+d^2)}=4+4+4=12$

In the following expression, the factors (x-2) cancel, so, our limit is ...

  $\displaystyle\lim_{x\to2}\dfrac{a-b}{c-d}=\lim_{x\to2}\dfrac{(a^2-b^2)(c^2+cd+d^2)}{(c^3-d^3)(a+b)}=\lim_{x\to2}\dfrac{-17(x-2)(12)}{-23(x-2)(6)}=\boxed{\dfrac{34}{23}}$