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Limit of the function
:
Limit of 7^(1/(-3+x))
Limit of (3-3*x^2+4*x^4+6*x^3)/(2*x^2+7*x^4)
Limit of ((5+4*x)/(-1+5*x))^(1+3*x)
Limit of (-6-x^2-3*x+4*x^3)/(3-x^2+2*x^3)
Derivative of
:
e^3*x^2
Identical expressions
e^ three *x^ two
e cubed multiply by x squared
e to the power of three multiply by x to the power of two
e3*x2
e³*x²
e to the power of 3*x to the power of 2
e^3x^2
e3x2
Limit of the function
/
e^3*x^2
Limit of the function e^3*x^2
at
→
Calculate the limit!
v
For end points:
---------
From the left (x0-)
From the right (x0+)
The graph:
from
to
Piecewise:
{
enter the piecewise function here
The solution
You have entered
[src]
/ 3 2\ lim \E *x / x->-oo
$$\lim_{x \to -\infty}\left(e^{3} x^{2}\right)$$
Limit(E^3*x^2, x, -oo)
Detail solution
Let's take the limit
$$\lim_{x \to -\infty}\left(e^{3} x^{2}\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to -\infty}\left(e^{3} x^{2}\right)$$ =
$$\lim_{x \to -\infty} \frac{1}{\frac{1}{x^{2}} e^{-3}}$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to -\infty} \frac{1}{\frac{1}{x^{2}} e^{-3}} = \lim_{u \to 0^+}\left(\frac{e^{3}}{u^{2}}\right)$$
=
$$\frac{e^{3}}{0} = \infty$$
The final answer:
$$\lim_{x \to -\infty}\left(e^{3} x^{2}\right) = \infty$$
Lopital's rule
There is no sense to apply Lopital's rule to this function since there is no indeterminateness of 0/0 or oo/oo type
The graph
Plot the graph
Rapid solution
[src]
oo
$$\infty$$
Expand and simplify
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to -\infty}\left(e^{3} x^{2}\right) = \infty$$
$$\lim_{x \to \infty}\left(e^{3} x^{2}\right) = \infty$$
More at x→oo
$$\lim_{x \to 0^-}\left(e^{3} x^{2}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(e^{3} x^{2}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(e^{3} x^{2}\right) = e^{3}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(e^{3} x^{2}\right) = e^{3}$$
More at x→1 from the right
The graph