## Evaluating Limits

Subscribe
I'd like to be

Inline Feedbacks
Chris
6 months ago

lim 3x/x-5 Hello! How do you solve this please? I cannot factor 3x.
x→5

Matheno
Editor
6 months ago

Thanks for asking! You’re of course correct that you can’t factor this expression — and as it turns out, the limit does not exist, as you can see by looking at the attached graph.

We discuss this situation above, right after Substitution Practice #3, and have the statement that is you try Substitution and find
$\frac{\text{a non-zero number}}{0}\text{,}$
then the limit does not exist.

Specifically in this case, when we try Substitution we get
$\lim_{x \to 5}\frac{3x}{x-5} = \frac{15}{0}$
This tells you that the numerator is approaching 15, while the denominator is getting closer and closer to zero, which means that the fraction “blows up” as x –> 5. Hence the limit does not exist.

We hope that helps, and please let us know if you have other questions!

1 year ago

$$\lim_{x \to 0}\left( 5x^5+4x^4-3x^3+2x^2-x+1\right)^3$$
how can you solve this I don’t get it.

Matheno
Editor
1 year ago

Thanks for asking! This is a problem that can use Tactic #1 above, Substitition. Remember, if you simply substitute in the value and get an answer that’s not 0/0, then you’re done. That works here:

\begin{align*}
\lim_{x \to 0}\left( 5x^5+4x^4-3x^3+2x^2-x+1\right)^3 &= \left( 5(0)^5 + 4(0)^4 -3(0)^3 + 2(0)^2 – 0 + 1\right)^3 \8px] &= (0 + 0 – 0 + 0 -0 + 1)^3 \\[8px] &= (1)^3 = 1 \quad \cmark \end{align*} We hope that helps. Please let us know if you have other questions! Nana 1 year ago If (1+x)^1/x is equal to e when lim x approaches to 0, (1+x^2)^1/x^2, is it equal to e^2? Matheno Editor Reply to Nana 1 year ago Hi, Nana, Thanks for asking! We’re going to provide two answers here for reasons you’ll see. [You may need to reload this page for the math below to render correctly.] First, the question doesn’t make sense to us since \displaystyle{\lim_{x \to 0} \dfrac{1+x}{x} \ne e }. Instead, since \[\frac{1+x}{x} = \frac{1}{x} + 1
and $\dfrac{1}{x}$ blows up as $x \rightarrow 0,$ the limit does not exist. The graph below also shows visually what happens near $x=0.$
$\lim_{x \to 0} \left(\dfrac{1}{x} + 1 \right) \text{ does not exist} \quad \cmark$
Since the “if” in your question isn’t correct, we can’t procede past that.

At the same time, we know that math can be difficult to type into a comment, and so you may not have written the equation you meant to. So let’s provide a more general answer to what you might be asking:

If $\displaystyle{\lim_{x \to 0} f(x) = L },$ does $\displaystyle{\lim_{x \to 0} [f(x)]^2 = L^2 }?$
Then the answer is yes. In general,
In general, $\displaystyle{\lim_{x \to a} [f(x)]^n = \left[\lim_{x \to a}f(x) \right]^n}$
where $n$ is a positive integer.

Note, however, that your second function as-written is not the square of the first function. If we take $f(x) = \dfrac{1+x}{x},$ then
$[f(x)]^2 = \dfrac{(1+x)^2}{x^2} = \dfrac{1 + 2x + x^2}{x^2} \ne \dfrac{1+x^2}{x^2}$
and so we can’t draw any “if … then …” conclusions.

If you’d like to clarify the question, we’d be happy to try to provide a more useful answer. As it is, we hope some of this is helpful!

Thanks again for asking. : )

Last edited 1 year ago by Matheno
Khalid khan
2 years ago

Lim (x^2+3x-10)/x^2+5
x=-5

Matheno
Editor
2 years ago

Thanks for asking, Khalid! (You may need to refresh your browser to see the answer below correctly rendered.)

Simple substitution works in this case, since the denominator is non-zero:
\begin{align*}
\lim_{x \to -5}\frac{x^2 +3x -10}{x^2 + 5} &= \frac{(-5)^2 + 3(-5) – 10}{(-5)^2+5} \\[8px]
&= \frac{25 – 15 -10}{25 + 5} \\[8px]
&= \frac{0}{30} \\[8px]