Question

lim x➡3
x^3 - 4x – 15\x^3 + x² - 6x – 18 ​

Answer 1

The vaue is "$\frac{23}{27}$".

Step-by-step explanation:

We have

$\lim_{x \to \3} \frac{x^{3} - 4\times\x - 15 }{x^{3} + x^{2} - 6\times\x - 18}$

Pux x = 0, It is$\dfrac{0}{0} fron.$

Applying L'Hospital rule,

Put  x = 3, we get

$\frac{3\times\3^{2} - 4 - 0}{3\times\3^{2}  + 2\times\3  -  6 - 0]</p><p>= [tex]\dfrac{27 - 4}{27 + 6 - 6}$

= $\dfrac{23}{27}$

Hence, the vaue is $\dfrac{23}{27}$.

Answer 2

The value of given limit is $\frac{23}{27}$.

Step-by-step explanation:

The given limit problem is

$lim_{x\rightarrow 3} (\dfrac{x^3 - 4x -15}{x^3 + x^2- 6x- 18})$

Apply limits.

$\dfrac{(3)^3 - 4(3) -15}{(3)^3 + (3)^2- 6(3)- 18})=\dfrac{0}{0}$

It is a $\frac{0}{0}$ form.

L'Hospital rule: After applying the limits if we get $\frac{0}{0}$ form, then

$lim_{x\rightarrow a}=\dfrac{f(x)}{g(x)}=lim_{x\rightarrow a}=\dfrac{f'(x)}{g'(x)}$

Using L'Hospital rule we get

$lim_{x\rightarrow 3} (\dfrac{x^3 - 4x -15}{x^3 + x^2- 6x- 18})=lim_{x\rightarrow 3} (\dfrac{\frac{d}{dx}(x^3 - 4x -15)}{\frac{d}{dx}(x^3 + x^2- 6x- 18)})$

$lim_{x\rightarrow 3} (\dfrac{x^3 - 4x -15}{x^3 + x^2- 6x- 18})=lim_{x\rightarrow 3} (\dfrac{3x^2 - 4}{3x^2 +2x- 6})$

Apply limits on right side.

$lim_{x\rightarrow 3} (\dfrac{x^3 - 4x -15}{x^3 + x^2- 6x- 18})=\dfrac{3(3)^2 - 4}{3(3)^2 +2(3)- 6}$

$lim_{x\rightarrow 3} (\dfrac{x^3 - 4x -15}{x^3 + x^2- 6x- 18})=\dfrac{23}{27}$

Therefore, the value of given limit is $\frac{23}{27}$.

#Learn more

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