Respuesta :
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You can interpret this question as one of the following:
• Interpretation I: Compute the limit:
[tex]\mathsf{\underset{x\to 1}{\ell im}~\dfrac{x^4-\sqrt{x}}{\sqrt{x}-1}\qquad\quad(i)}[/tex]
Make a substitution:
[tex]\begin{array}{lcl} \mathsf{\sqrt{x}=u}&\quad\Rightarrow\quad&\mathsf{x=u^2}\\\\ &&\mathsf{x^4=(u^2)^4}\\\\ &&\mathsf{x^4=u^8} \end{array}[/tex]
and u approaches 1 as x approaches 1. So, the limit (i) becomes
[tex]\mathsf{=\underset{u\to 1}{\ell im}~\dfrac{u^8-u}{u-1}\qquad\quad(ii)}[/tex]
You still have a "0/0" indeterminate form. So it's possible to factor out (u – 1) from the numerator, by polynomial division. Observe:
u⁸ – u
= u⁸ – u⁷ + u⁷ – u
= u⁷(u – 1) + u⁷ – u
= u⁷(u – 1) + u⁷ – u⁶ + u⁶ – u
= u⁷(u – 1) + u⁶(u – 1) + u⁶ – u
= u⁷(u – 1) + u⁶(u – 1) + u⁶ – u⁵ + u⁵ – u
= u⁷(u – 1) + u⁶(u – 1) + u⁵(u – 1) + u⁵ – u
= u⁷(u – 1) + u⁶(u – 1) + u⁵(u – 1) + u⁵ – u⁴ + u⁴ – u
= u⁷(u – 1) + u⁶(u – 1) + u⁵(u – 1) + u⁴(u – 1) + u⁴ – u
= u⁷(u – 1) + u⁶(u – 1) + u⁵(u – 1) + u⁴(u – 1) + u⁴ – u³ + u³ – u
= u⁷(u – 1) + u⁶(u – 1) + u⁵(u – 1) + u⁴(u – 1) + u³(u – 1) + u³ – u
= u⁷(u – 1) + u⁶(u – 1) + u⁵(u – 1) + u⁴(u – 1) + u³(u – 1) + u³ – u² + u² – u
= u⁷(u – 1) + u⁶(u – 1) + u⁵(u – 1) + u⁴(u – 1) + u³(u – 1) + u²(u – 1) + u(u – 1)
Take out (u – 1) as a common factor and you get
u⁸ – u = (u⁷ + u⁶ + u⁵ + u⁴ + u³ + u² + u) · (u – 1)
Substitute for u⁸ – u in the numerator, and then simplify the common factors, so the limit (ii) becomes
[tex]\mathsf{=\underset{u\to 1}{\ell im}~\dfrac{(u^7+u^6+u^5+u^4+u^3+u^2+u)\cdot (u-1)}{u-1}}\\\\\\ \mathsf{=\underset{u\to 1}{\ell im}~(u^7+u^6+u^5+u^4+u^3+u^2+u)}\\\\\\ \mathsf{=1^7+1^6+1^5+1^4+1^3+1^2+1}\\\\ \mathsf{=1+1+1+1+1+1+1}[/tex]
= 7 ✔
[tex]\boxed{\begin{array}{c} \mathsf{\underset{x\to 1}{\ell im}~\dfrac{x^4-\sqrt{x}}{\sqrt{x}-1}=7} \end{array}}[/tex] ✔
________
• Interpretation II: Compute the limit:
[tex]\mathsf{\underset{x\to 1}{\ell im}~\dfrac{x^4-\sqrt{x}}{\sqrt{x-1}}\qquad\quad(iii)}[/tex]
Multiply both the numerator and denominator by the numerator's conjugate, which is [tex]\mathsf{(x^4+\sqrt{x}):}[/tex]
[tex]\mathsf{=\underset{x\to 1}{\ell im}~\dfrac{(x^4-\sqrt{x})\cdot (x^4+\sqrt{x})}{\sqrt{x-1}\cdot (x^4+\sqrt{x})}}\\\\\\ \mathsf{=\underset{x\to 1}{\ell im}~\dfrac{(x^4)^2-(\sqrt{x})^2}{\sqrt{x-1}\cdot (x^4+\sqrt{x})}}\\\\\\ \mathsf{=\underset{x\to 1}{\ell im}~\dfrac{x^8-x}{\sqrt{x-1}\cdot (x^4+\sqrt{x})}}[/tex]
Now, multiply again both the numerator and denominator by [tex]\mathsf{\sqrt{x-1},}[/tex] so you can simplify that square root:
[tex]\mathsf{=\underset{x\to 1}{\ell im}~\dfrac{(x^8-x)\cdot \sqrt{x-1}}{\sqrt{x-1}\cdot (x^4+\sqrt{x})\cdot \sqrt{x-1}}}\\\\\\ \mathsf{=\underset{x\to 1}{\ell im}~\dfrac{(x^8-x)\cdot \sqrt{x-1}}{(\sqrt{x-1})^2\cdot (x^4+\sqrt{x})}}\\\\\\ \mathsf{=\underset{x\to 1}{\ell im}~\dfrac{(x^8-x)\cdot \sqrt{x-1}}{(x-1)\cdot (x^4+\sqrt{x})}}\\\\\\ \mathsf{=\underset{x\to 1}{\ell im}~\dfrac{x^8-x}{x-1}\cdot \dfrac{\sqrt{x-1}}{x^4+\sqrt{x}}\qquad\quad(iv)}[/tex]
Factoring out (x – 1) from the numerator in the first fraction, as we did in interpretation I, you get
x⁸ – x = (x⁷ + x⁶ + x⁵ + x⁴ + x³ + x² + x) · (x – 1)
Substitute for x⁸ – x in the numerator, and then simplify the common factors, so the limit (iv) becomes
[tex]\mathsf{=\underset{x\to 1}{\ell im}~\dfrac{(x^7+x^6+x^5+x^4+x^3+x^2+x)\cdot (x-1)}{x-1}\cdot \dfrac{\sqrt{x-1}}{x^4+\sqrt{x}}}\\\\\\ \mathsf{=\underset{x\to 1}{\ell im}~(x^7+x^6+x^5+x^4+x^3+x^2+x)\cdot \dfrac{\sqrt{x-1}}{x^4+\sqrt{x}}}\\\\\\ \mathsf{=(1^7+1^6+1^5+1^4+1^3+1^2+1)\cdot \dfrac{\sqrt{1-1}}{1^4+\sqrt{1}}}\\\\\\ \mathsf{=(1+1+1+1+1+1+1)\cdot \dfrac{\sqrt{0}}{1+1}}\\\\\\ \mathsf{=7\cdot \dfrac{0}{2}}[/tex]
= 0 ✔
[tex]\boxed{\begin{array}{c} \mathsf{\underset{x\to 1}{\ell im}~\dfrac{x^4-\sqrt{x}}{\sqrt{x-1}}=0} \end{array}}[/tex] ✔
And now, just choose one of those interpretations that apply. =)
I hope this helps. =)
Tags: limit square root sqrt irrational substitution rational conjugate simplify differential integral calculus
______________
You can interpret this question as one of the following:
• Interpretation I: Compute the limit:
[tex]\mathsf{\underset{x\to 1}{\ell im}~\dfrac{x^4-\sqrt{x}}{\sqrt{x}-1}\qquad\quad(i)}[/tex]
Make a substitution:
[tex]\begin{array}{lcl} \mathsf{\sqrt{x}=u}&\quad\Rightarrow\quad&\mathsf{x=u^2}\\\\ &&\mathsf{x^4=(u^2)^4}\\\\ &&\mathsf{x^4=u^8} \end{array}[/tex]
and u approaches 1 as x approaches 1. So, the limit (i) becomes
[tex]\mathsf{=\underset{u\to 1}{\ell im}~\dfrac{u^8-u}{u-1}\qquad\quad(ii)}[/tex]
You still have a "0/0" indeterminate form. So it's possible to factor out (u – 1) from the numerator, by polynomial division. Observe:
u⁸ – u
= u⁸ – u⁷ + u⁷ – u
= u⁷(u – 1) + u⁷ – u
= u⁷(u – 1) + u⁷ – u⁶ + u⁶ – u
= u⁷(u – 1) + u⁶(u – 1) + u⁶ – u
= u⁷(u – 1) + u⁶(u – 1) + u⁶ – u⁵ + u⁵ – u
= u⁷(u – 1) + u⁶(u – 1) + u⁵(u – 1) + u⁵ – u
= u⁷(u – 1) + u⁶(u – 1) + u⁵(u – 1) + u⁵ – u⁴ + u⁴ – u
= u⁷(u – 1) + u⁶(u – 1) + u⁵(u – 1) + u⁴(u – 1) + u⁴ – u
= u⁷(u – 1) + u⁶(u – 1) + u⁵(u – 1) + u⁴(u – 1) + u⁴ – u³ + u³ – u
= u⁷(u – 1) + u⁶(u – 1) + u⁵(u – 1) + u⁴(u – 1) + u³(u – 1) + u³ – u
= u⁷(u – 1) + u⁶(u – 1) + u⁵(u – 1) + u⁴(u – 1) + u³(u – 1) + u³ – u² + u² – u
= u⁷(u – 1) + u⁶(u – 1) + u⁵(u – 1) + u⁴(u – 1) + u³(u – 1) + u²(u – 1) + u(u – 1)
Take out (u – 1) as a common factor and you get
u⁸ – u = (u⁷ + u⁶ + u⁵ + u⁴ + u³ + u² + u) · (u – 1)
Substitute for u⁸ – u in the numerator, and then simplify the common factors, so the limit (ii) becomes
[tex]\mathsf{=\underset{u\to 1}{\ell im}~\dfrac{(u^7+u^6+u^5+u^4+u^3+u^2+u)\cdot (u-1)}{u-1}}\\\\\\ \mathsf{=\underset{u\to 1}{\ell im}~(u^7+u^6+u^5+u^4+u^3+u^2+u)}\\\\\\ \mathsf{=1^7+1^6+1^5+1^4+1^3+1^2+1}\\\\ \mathsf{=1+1+1+1+1+1+1}[/tex]
= 7 ✔
[tex]\boxed{\begin{array}{c} \mathsf{\underset{x\to 1}{\ell im}~\dfrac{x^4-\sqrt{x}}{\sqrt{x}-1}=7} \end{array}}[/tex] ✔
________
• Interpretation II: Compute the limit:
[tex]\mathsf{\underset{x\to 1}{\ell im}~\dfrac{x^4-\sqrt{x}}{\sqrt{x-1}}\qquad\quad(iii)}[/tex]
Multiply both the numerator and denominator by the numerator's conjugate, which is [tex]\mathsf{(x^4+\sqrt{x}):}[/tex]
[tex]\mathsf{=\underset{x\to 1}{\ell im}~\dfrac{(x^4-\sqrt{x})\cdot (x^4+\sqrt{x})}{\sqrt{x-1}\cdot (x^4+\sqrt{x})}}\\\\\\ \mathsf{=\underset{x\to 1}{\ell im}~\dfrac{(x^4)^2-(\sqrt{x})^2}{\sqrt{x-1}\cdot (x^4+\sqrt{x})}}\\\\\\ \mathsf{=\underset{x\to 1}{\ell im}~\dfrac{x^8-x}{\sqrt{x-1}\cdot (x^4+\sqrt{x})}}[/tex]
Now, multiply again both the numerator and denominator by [tex]\mathsf{\sqrt{x-1},}[/tex] so you can simplify that square root:
[tex]\mathsf{=\underset{x\to 1}{\ell im}~\dfrac{(x^8-x)\cdot \sqrt{x-1}}{\sqrt{x-1}\cdot (x^4+\sqrt{x})\cdot \sqrt{x-1}}}\\\\\\ \mathsf{=\underset{x\to 1}{\ell im}~\dfrac{(x^8-x)\cdot \sqrt{x-1}}{(\sqrt{x-1})^2\cdot (x^4+\sqrt{x})}}\\\\\\ \mathsf{=\underset{x\to 1}{\ell im}~\dfrac{(x^8-x)\cdot \sqrt{x-1}}{(x-1)\cdot (x^4+\sqrt{x})}}\\\\\\ \mathsf{=\underset{x\to 1}{\ell im}~\dfrac{x^8-x}{x-1}\cdot \dfrac{\sqrt{x-1}}{x^4+\sqrt{x}}\qquad\quad(iv)}[/tex]
Factoring out (x – 1) from the numerator in the first fraction, as we did in interpretation I, you get
x⁸ – x = (x⁷ + x⁶ + x⁵ + x⁴ + x³ + x² + x) · (x – 1)
Substitute for x⁸ – x in the numerator, and then simplify the common factors, so the limit (iv) becomes
[tex]\mathsf{=\underset{x\to 1}{\ell im}~\dfrac{(x^7+x^6+x^5+x^4+x^3+x^2+x)\cdot (x-1)}{x-1}\cdot \dfrac{\sqrt{x-1}}{x^4+\sqrt{x}}}\\\\\\ \mathsf{=\underset{x\to 1}{\ell im}~(x^7+x^6+x^5+x^4+x^3+x^2+x)\cdot \dfrac{\sqrt{x-1}}{x^4+\sqrt{x}}}\\\\\\ \mathsf{=(1^7+1^6+1^5+1^4+1^3+1^2+1)\cdot \dfrac{\sqrt{1-1}}{1^4+\sqrt{1}}}\\\\\\ \mathsf{=(1+1+1+1+1+1+1)\cdot \dfrac{\sqrt{0}}{1+1}}\\\\\\ \mathsf{=7\cdot \dfrac{0}{2}}[/tex]
= 0 ✔
[tex]\boxed{\begin{array}{c} \mathsf{\underset{x\to 1}{\ell im}~\dfrac{x^4-\sqrt{x}}{\sqrt{x-1}}=0} \end{array}}[/tex] ✔
And now, just choose one of those interpretations that apply. =)
I hope this helps. =)
Tags: limit square root sqrt irrational substitution rational conjugate simplify differential integral calculus
