Question

need answers...
topic: LIMITS
1. lim x3-9x2+25x-20/x3-4x2-4x+16
x→4
2. lim 3x3-x2-7x+5/2x3-x2-4x+3
x→ 1
3. lim 2x3-x2+5/4x3+1
x→∞
4. lim 4x +5/√9x2-1
x→∞
5. lim x2-5x/sinx
x→ 0

Answer 1

@ganeshie8

Answer 2

@mathmale

Answer 3

Hello, Christine,

I have some suggestions to help you get started. Please remember that the goal of OpenStudy is not to provide you with answers, but rather with guidance that will hopefully enable you to find your own answers.

In regard to your first problem:

1. lim x3-9x2+25x-20/x3-4x2-4x+16
x→4

This needs re-writing, with parentheses, because otherwise there is ambiguity in regard to what quantity divides into what. The simplest way to remove that ambiguity follows:

1. lim (x3-9x2+25x-20)/(x3-4x2-4x+16)
x→4

Note how that re-written limit clearly shows that you are dividing all four terms of the numerator by all 4 terms of the denominator. This is critically important.

Next: Determine what happens to the numerator as x approaches 4 (is that what yuo meant?). If the denom. approaches zero, we've got a problem and are going to have to factor both numerator and denominator and determine whether there are any common factors that could be cancelled.

Alternatively, you could start immediately by factoring both the numerator and the den. and cancelling out any common factors.

Please get started. Show your work here, so that others can give you relevant feedback regarding what you're doing.

Answer 4

Also, for the sake of clarity, would you please express exponents properly:

x3-9x2+25x-20/x3-4x2-4x+16 should appear as

(x^3 - 9x^2 - 25x - 20) / (x^3 - 4x^2 -4x +16).

x3: looks like \[x _{3}\]
x^3 looks like and means \[x ^{3}\]

Answer 5

@abb0t @beccamarx19 @bquiz

Answer 6

@Tushara

Answer 7

\[\begin{align*}\lim_{x\to4}\frac{x^3-9x^2+25x-20}{x^3-4x^2-4x+16}&=\lim_{x\to4}\frac{\color{red}{(x-4)(x^2-5x+5)}}{x^2(x-4)-4(x-4)}\\
&=\lim_{x\to4}\frac{x^2-5x+5}{x^2-4}
\end{align*}\]
For the red numerator: you want to eliminate the discontinuity at \(x=4\), so you divide \(x^3-9x^2+25x-20\) by \(x-4\). Long division or synthetic division should help you there.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
\[\begin{align*}\lim_{x\to1}\frac{3x^3-x^2-7x+5}{2x^3-x^2-4x+3}&=\lim_{x\to1}\frac{(x-1)(3x^2+2x-5)}{(x-1)(2x^2+x-3)}\\
&=\lim_{x\to1}\frac{3x^2+2x-5}{2x^2+x-3}
\end{align*}\]
Same strategy. Factorization.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
\[\begin{align*}\lim_{x\to\infty}\frac{2x^3-x^2+5}{4x^3+1}&=\lim_{x\to\infty}\frac{x^3\left(2-\dfrac{1}{x}+\dfrac{5}{x^3}\right)}{x^3\left(4+\dfrac{1}{x^3}\right)}\\
&=\lim_{x\to\infty}\frac{2-\dfrac{1}{x}+\dfrac{5}{x^3}}{4+\dfrac{1}{x^3}}\\
\end{align*}\]
More factorization. This time, pull out the highest power of \(x\).
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
\[\begin{align*}\lim_{x\to\infty}\frac{4x+5}{\sqrt{9x^2-1}}&=\lim_{x\to\infty}\frac{x\left(4+\dfrac{5}{x}\right)}{\sqrt{x^2}\sqrt{9-\dfrac{1}{x^2}}}\\
&=\lim_{x\to\infty}\frac{x\left(4+\dfrac{5}{x}\right)}{|x|\sqrt{9-\dfrac{1}{x^2}}}\\
&=\lim_{x\to\infty}\frac{x\left(4+\dfrac{5}{x}\right)}{x\sqrt{9-\dfrac{1}{x^2}}}\\
&=\lim_{x\to\infty}\frac{4+\dfrac{5}{x}}{\sqrt{9-\dfrac{1}{x^2}}}\end{align*}\]
\(|x|=x\) for positive values of \(x\), which is the case because \(x\to\infty\).
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
\[\begin{align*}\lim_{x\to0}\frac{x^2-4x}{\sin x}&=\lim_{x\to0}\frac{x(x-4)}{\sin x}\\
&=\left(\lim_{x\to0}\frac{x}{\sin x}\right)\cdot\left(\lim_{x\to0}(x-4)\right)\\
&=\lim_{x\to0}(x-4)
\end{align*}\]
Recall the limit, \(\displaystyle\lim_{x\to0}\frac{\sin ax}{ax}=\lim_{x\to0}\frac{ax}{\sin ax}=1\) for \(a\not=0\).