#  ## Evaluating Limits

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Nana
2 months 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
2 months 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 2 months ago by Matheno
Khalid khan
5 months ago

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

Matheno
Editor
5 months 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]