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Limit of the function
:
Limit of ((1+x)^4-(-1+x)^4)/((1+x)^4+(-1+x)^4)
Limit of (-2*sin(x)+sin(2*x))/(x*log(cos(5*x)))
Limit of (3-4*x^2+8*x^4)/(1+2*x^4)
Limit of (8-7*x)/(-5-4*x+6*x^2)
Graphing y =
:
x/6
Derivative of
:
x/6
Identical expressions
x/ six
x divide by 6
x divide by six
Similar expressions
sin(4*x)/(6*x)
(-4+2*x)/(6+4*x+7*x^2)
sin(7*x)/(6*x)
x*sin(5*x)/6
(3+x)/(6+x)
Limit of the function
/
x/6
Limit of the function x/6
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]
/x\ lim |-| x->oo\6/
$$\lim_{x \to \infty}\left(\frac{x}{6}\right)$$
Limit(x/6, x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(\frac{x}{6}\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(\frac{x}{6}\right)$$ =
$$\lim_{x \to \infty} \frac{1}{6 \frac{1}{x}}$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty} \frac{1}{6 \frac{1}{x}} = \lim_{u \to 0^+}\left(\frac{1}{6 u}\right)$$
=
$$\frac{1}{0 \cdot 6} = \infty$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{x}{6}\right) = \infty$$
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
$$\lim_{x \to \infty}\left(\frac{x}{6}\right) = \infty$$
$$\lim_{x \to 0^-}\left(\frac{x}{6}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x}{6}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x}{6}\right) = \frac{1}{6}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x}{6}\right) = \frac{1}{6}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x}{6}\right) = -\infty$$
More at x→-oo
The graph