Please help me find limit

Mathematics

1 answer:

cluponka [151]2 years ago

4 0

9514 1404 393

Answer:

-13/11

Step-by-step explanation:

Straightforward evaluation of the expression at x=1 gives (1 -1)/(1 -1) = 0/0, an indeterminate form. So, L'Hopital's rule applies. The ratio of derivatives is ...

$\displaystyle\lim_{x\to 1}\dfrac{n}{d}=\dfrac{n'}{d'}=\left.\dfrac{\dfrac{4}{3\sqrt[3]{4x-3}}-\dfrac{7}{2\sqrt{7x-6}}}{\dfrac{5}{2\sqrt{5x-4}}-\dfrac{2}{3\sqrt[3]{2x-1}}}\right|_{x=1}=\dfrac{4/3-7/2}{5/2-2/3}=\dfrac{8-21}{15-4}\\\\=\boxed{-\dfrac{13}{11}}$

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The length of a rectangle is 2 centimeters less than its width. What are the dimensions of the rectangle if its area is 195 squa

JulijaS [17]

Answer:

Length=15 cm

Width=13 cm

Step-by-step explanation:

-Let x be the width dimension, the length will be (x-2)

-The area of a rectangle is given by the formula;

$A=lw$

#We substitute the length and width values and equate to the given area:

$A=lw\\195=x\times (x-2)\\\\195=x^2-2x$

#Rewrite as a quadratic and solve for x:

$195=x^2-2x\\\\x^2-2x-195=0\\\\\# Use \ the \ Quadratic\ formula \ to\ solve:\\\\x_1,x_2=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\therefore x_1,x_2=\frac{-(-2)\pm\sqrt{(-2)^2-4\times 1\times(-195)}}{2\times 1}\\\\\\x_1=15, x_2=-13$