Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
-
Limit of x^(4/x)
-
Limit of (-6+x^2-x)/(-4+x^2)
-
Limit of (-6+x^2-x)/(-21+x+2*x^2)
-
Limit of e^(3-x)*(-2+x)
- Graphing y =:
- 3+3*x
-
Identical expressions
- three + three *x
- 3 plus 3 multiply by x
- three plus three multiply by x
- 3+3x
-
Similar expressions
- 3-3*x
Limit of the function 3+3*x
The solution
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lim (3 + 3*x) x->oo
$$\lim_{x \to \infty}\left(3 x + 3\right)$$
Limit(3 + 3*x, x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(3 x + 3\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(3 x + 3\right)$$ =
$$\lim_{x \to \infty}\left(\frac{3 + \frac{3}{x}}{\frac{1}{x}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{3 + \frac{3}{x}}{\frac{1}{x}}\right) = \lim_{u \to 0^+}\left(\frac{3 u + 3}{u}\right)$$
=
$$\frac{0 \cdot 3 + 3}{0} = \infty$$
The final answer:
$$\lim_{x \to \infty}\left(3 x + 3\right) = \infty$$
$$\lim_{x \to \infty}\left(3 x + 3\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(3 x + 3\right)$$ =
$$\lim_{x \to \infty}\left(\frac{3 + \frac{3}{x}}{\frac{1}{x}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{3 + \frac{3}{x}}{\frac{1}{x}}\right) = \lim_{u \to 0^+}\left(\frac{3 u + 3}{u}\right)$$
=
$$\frac{0 \cdot 3 + 3}{0} = \infty$$
The final answer:
$$\lim_{x \to \infty}\left(3 x + 3\right) = \infty$$
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
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to \infty}\left(3 x + 3\right) = \infty$$
$$\lim_{x \to 0^-}\left(3 x + 3\right) = 3$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(3 x + 3\right) = 3$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(3 x + 3\right) = 6$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(3 x + 3\right) = 6$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(3 x + 3\right) = -\infty$$
More at x→-oo
$$\lim_{x \to 0^-}\left(3 x + 3\right) = 3$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(3 x + 3\right) = 3$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(3 x + 3\right) = 6$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(3 x + 3\right) = 6$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(3 x + 3\right) = -\infty$$
More at x→-oo
The graph