Limits

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Evaluate the limit:

$\lim_{x \to 3}\frac{x^2-x-6}{x^2-9}$

$\frac{5}{6}$

$\infty$

Explanation

The limiting situation in this equation would be the denominator. Plug the value that is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will equal zero when x=3; so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of x into the remaining equation.

$\lim_{x \to 3}\frac{x^2-x-6}{x^2-9}=\lim_{x \to 3}\frac{(x-3)(x+2)}{(x-3)(x+3)}=\lim_{x \to 3}\frac{x+2}{x+3}=\frac{3+2}{3+3}=\frac{5}{6}$

Evaluate the limit:

$\lim_{x \to 0}\frac{x^2+2x+1}{x+6}$

$\frac{1}{6}$

Does not exist

$-\infty$

$\infty$

Explanation

The limiting situation in this equation would be the denominator. Plug the value that x is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will not equal zero when x=0; so we proceed to insert the value of x into the entire equation.

$\lim_{x \to 0}\frac{x^2+2x+1}{x+6}=\frac{0^2+2*0+1}{0+6}=\frac{1}{6}$

$\begin{align*}&\text{Calculate the limits of the function}\&lim_{(x,y)\rightarrow(0-,0-)}-\frac{(2x^{3}y^{3} + x^{5}y - 3x^{6})}{(3x^{6} + y^{6})}\&\text{along the x-axis and the line, }y=x\end{align*}$

$\begin{align*}&\text{x-axis: }0\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\&y=x:-\frac{41}{8}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{50}{19}\&y=x:-\frac{39}{17}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{32}{13}\&y=x:\frac{50}{11}\end{align*}$

Explanation

$\begin{align*}&\text{To approach this problem, it is important to understand what}\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\&-\frac{(2(0)^{3}y^{3} + (0)^{5}y - 3(0)^{6})}{(3(0)^{6} + y^{6})}=\frac{0}{y^{6}}\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{y^{6}}=0\&\text{We can make a similar substitution for }y=x:\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(2x^{3}(x)^{3} + x^{5}(x) - 3x^{6})}{(3x^{6} + (x)^{6})}=0\&\text{Since the two limits are equal, it's possible the limit exists.}\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

Evaluate the limit:

$\lim_{x \to 3}\frac{x^2-x-6}{x^2-9}$

$\frac{5}{6}$

$\infty$

Explanation

$\lim_{x \to 3}\frac{x^2-x-6}{x^2-9}=\lim_{x \to 3}\frac{(x-3)(x+2)}{(x-3)(x+3)}=\lim_{x \to 3}\frac{x+2}{x+3}=\frac{3+2}{3+3}=\frac{5}{6}$

Evaluatle the limit:

$\lim_{x \to 3} x^2-2x+6$

Explanation

Consider the domain of the function. Because this equation is a polynomial, x is not restricted by any value. Thus the way to evaluate this limit would simply be to plug the value that x is approaching into the limit equation.

$\lim_{x \to 3} x^2-2x+6=3^2-2*3+6=9-6+6=9$

Evaluate the limit:

$\lim_{x \to 0}\frac{x^2+2x+1}{x+6}$

$\frac{1}{6}$

Does not exist

$-\infty$

$\infty$

Explanation

$\lim_{x \to 0}\frac{x^2+2x+1}{x+6}=\frac{0^2+2*0+1}{0+6}=\frac{1}{6}$

$\begin{align*}&\text{Evaluate the following limit:}\&lim_{x\rightarrow0-}\frac{9sin(6x)^{2}}{7x^{2}}\end{align*}$

$\frac{324}{7}$

$\infty$

$-\infty$

$-\frac{324}{7}$

Explanation

$\begin{align*}&\text{Examining this problem, we find that both}\&\text{numerator and denominator have zero values at the limit specified.}\&\text{Noting this, L'Hopital's method may be of use:}\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\&\frac{9sin(6x)^{2}}{7x^{2}}\rightarrow\frac{0}{0}\&\frac{108cos(6x)sin(6x)}{14x}\rightarrow\frac{0}{0}\&\frac{648cos(6x)^{2} - 648sin(6x)^{2}}{14}\rightarrow\frac{648}{14}\&lim_{x\rightarrow0-}\frac{9sin(6x)^{2}}{7x^{2}}=\frac{324}{7}\end{align*}$

$\begin{align*}&\text{Evaluate the following limit:}\&lim_{x\rightarrow1-}\frac{-4\cdot (x - 1)^{2}}{9sin(2\cdot \pi x)^{2}}\end{align*}$

$-\frac{1}{(9\cdot \pi ^{2})}$

$\frac{1}{(9\cdot \pi ^{2})}$

$\infty$

$-\infty$

Explanation

$\begin{align*}&\text{Examining this problem, we find that both}\&\text{numerator and denominator have zero values at the limit specified.}\&\text{Noting this, L'Hopital's method may be of use:}\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\&\frac{-4\cdot (x - 1)^{2}}{9sin(2\cdot \pi x)^{2}}\rightarrow\frac{0}{0}\&\frac{8 - 8x}{36\cdot \pi cos(2\cdot \pi x)sin(2\cdot \pi x)}\rightarrow\frac{0}{0}\&\frac{-8}{72\cdot \pi ^{2}cos(2\cdot \pi x)^{2} - 72\cdot \pi ^{2}sin(2\cdot \pi x)^{2}}\rightarrow\frac{-8}{72\cdot \pi ^{2}}\&lim_{x\rightarrow1-}\frac{-4\cdot (x - 1)^{2}}{9sin(2\cdot \pi x)^{2}}=-\frac{1}{(9\cdot \pi ^{2})}\end{align*}$

If it exists, find the following limit

$\small \small \lim_{x\to 2} \frac{1}{x+2}$

$\small \frac{1}{4}$

$\small \frac{1}{2}$

$\small -\frac{1}{4}$

Does not exist

Explanation

To find the following limit

$\small \small \lim_{x\to 2} \frac{1}{x+2}$

we can simply plug in $\small x=2$ because it's defined at that point and in infinitesimally small neighborhoods near $\small x=2$ , so we get

$\small \small \small \lim_{x\to 2} \frac{1}{x+2}=\frac{1}{2+2}=\frac{1}{4}$

Screen shot 2015 07 22 at 11.03.39 am

Given the above graph of , what is $\lim_{x\rightarrow 0^{+}}f(x)$ ?

$\infty$

$-\infty$

Explanation

Examining the graph, we want to find where the graph tends to as it approaches zero from the right hand side. We can see that there appears to be a vertical asymptote at zero. As the x values approach zero from the right, the function values of the graph tend towards positive infinity.

Therefore, we can observe that $\lim_{x\rightarrow 0^{+}}f(x)=\infty$ as approaches from the right.

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