Question

Problem 1. Determine the following limits, or explain why they do not exist:

Answer 1

what limits?

Answer 2

I only need help with m

Answer 3

so for a fraction to be defined the denominator not can be equal never zero

in these cases how i see the denominator come to be equal zero - in this way these fractions will be undefined

Answer 4

Help please

Answer 5

For each problem, as the limit approaches a number, find what y-value is being approached from both sides, for example if the limit approaches h=2
then find the values at like 1.9 and 1.99 and see what they approach and similarly for the other side

Answer 6

I don't get it I am confused

Answer 7

hmm ok i have got a bit time

so were given this:
\[ \lim _{x \rightarrow \infty}\left[\frac{\sqrt{x^2-4}-\sqrt{x^2+27}}{x}\right] \]
pay attention to EVERY step.
\[ \lim _{x \rightarrow \infty}\left[\frac{\sqrt{x^2-4}-\sqrt{\color{lightskyblue}{x}^2+27}}{\color{lightskyblue}{x}}\right] \]
\[ \lim _{x \rightarrow \infty} \frac{\sqrt{\frac{x^2-4}{x^2}}-\sqrt{\frac{x^2}{x^2}+\frac{27}{x^2}}} {1} \Rightarrow \lim _{x \rightarrow \infty} \frac{\sqrt{\frac{x^2-4}{x^2}}-\sqrt{\frac{\cancel{x^2}}{\cancel{x^2}}+\frac{27}{x^2}}} {1}\]
\[ \lim _{x \rightarrow \infty} \frac{\sqrt{(1\times){\frac{x^2-4}{x^2}}}-(1\times)\sqrt{1+\frac{27}{x^2}}} {1} \]
\[ \frac{\sqrt{ \lim _{x \rightarrow \infty}{\frac{\cancel{x^2}-4}{\cancel{x^2}}}}- \lim _{x \rightarrow \infty}\sqrt{1+\frac{27}{x^2}}} { \lim _{x \rightarrow \infty}1} \]

\[ \frac{\sqrt{ \lim _{x \rightarrow \infty}{\frac{1-\frac{4}{x^2}}{1}}}- \lim _{x \rightarrow \infty}\sqrt{1+\frac{27}{x^2}}} { \lim _{x \rightarrow \infty}1} \]
\[ \frac{\sqrt{ {\frac{\lim _{x \rightarrow \infty}1-\lim _{x \rightarrow \infty}\frac{4}{x^2}}{\lim _{x \rightarrow \infty}1}}}- \lim _{x \rightarrow \infty}\sqrt{1+\frac{27}{x^2}}} { \lim _{x \rightarrow \infty}1} \]
\[ \frac{\sqrt{{\frac{1-4 \ \lim _{x \rightarrow \infty}\frac{1}{x^2}}{1}}}- \lim _{x \rightarrow \infty}\sqrt{1+\frac{27}{x^2}}} { \lim _{x \rightarrow \infty}1} \] since the numerator is a real number whilst its denominator is just unbounded, that means it approaches 0
\[ \frac{\sqrt{{\frac{ 1-4 \times 0}{ \lim _{x \rightarrow \infty}1}}}- \lim _{x \rightarrow \infty}\sqrt{1+\frac{27}{x^2}}} { \lim _{x \rightarrow \infty}1} \]
\[ \frac{\sqrt{{\frac{ 1-4 \times 0}{ \lim _{x \rightarrow \infty}1}}}- \sqrt{\lim _{x \rightarrow \infty}1+\frac{27}{x^2}}} { \lim _{x \rightarrow \infty}1} \]
do the same thing for the other side as well -- kinda lazy to do the latex, but you will end up with:
\[ \frac{\sqrt{{\frac{ 1-4 \times 0}{ \lim _{x \rightarrow \infty}1}}}- \sqrt{1+27\times 0 }} { \lim _{x \rightarrow \infty}1} \]
evaluate the limits and then get rid of the fraction form:
\[ \sqrt{{ 1-4 \times 0}}- \sqrt{1+27\times 0 } \]
\[ \frac{\sqrt{{\frac{ 1-4 \times 0}{\cancel1}}}- \sqrt{1+27\times 0 }} {\cancel 1} \]
\[ \sqrt{{ 1-4 \times 0}}- \sqrt{1+27\times 0 } \ \ \ \ \ \ \ \ \ \ \text{its all times 0} \]
\[ 1-1 \]
\[ 0 \]
thus, it doesnt exist. yay.

Answer 8

Flor did an excellent job explaining how to calculate the limit

But a point of contention, if I may.

Limit approaching 0 \(\neq\) Limit DNE

The limit will not exist at a point if you get two different values when you approach from the left and right side

A simple example
\[\lim_{x \rightarrow 0} \dfrac{1}{x} = \text{DNE}\]
You can see that as you approach from the left, as x becomes -0.1 then -0.01, -0.00001 and closer and closer to 0
The limit approaches -\(\infty\)
As you approach 1/x from the right, so as x becomes 0.1 then 0.01 and 0.00001 and so on closer to 0
The limit approaches \(\infty\)

Now take a look at
\[\lim_{x \rightarrow \infty} \dfrac{1}{x} = 0\]
When you increase the x value to something very large
1/100 = 0.01
1/1000000 = 0.000001
1/10000000000 = a number even smaller

and so as x gets closer to \(\infty\) the limit is approaching 0
And you can take a look at a graph to see this too

Answer 9

yes, i see where youre going, but what we have is the rational function of \( \lim_{x \rightarrow \infty} \dfrac{1}{x^{\color{lightskyblue}2}} ​\)

as you may know its bottom heavy so automatically its 0 yes?

thats why we say that \( \lim_{x \rightarrow \infty} \dfrac{1}{x^{2}} ​\) is \(=0\)

Answer 10

yes, I agree

it was just that at the end of your previous reply, you said
"0 thus, it doesnt exist"

so I wanted to clarify that just in case it wasn't clear :)