Mister Exam
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How to use it?
- Limit of the function:
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Limit of (-5+7*n)/(2+n)
-
Limit of (-64+4^x)/(-3+x)
-
Limit of (-7+3*x^2+5*x)/(1+x+3*x^2)
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Limit of (1+6*x)^(1/x)
- Graphing y =:
- 3+4*x
-
Identical expressions
- three + four *x
- 3 plus 4 multiply by x
- three plus four multiply by x
- 3+4x
-
Similar expressions
- 3-4*x
Limit of the function 3+4*x
The solution
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lim (3 + 4*x) x->oo
$$\lim_{x \to \infty}\left(4 x + 3\right)$$
Limit(3 + 4*x, x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(4 x + 3\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(4 x + 3\right)$$ =
$$\lim_{x \to \infty}\left(\frac{4 + \frac{3}{x}}{\frac{1}{x}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{4 + \frac{3}{x}}{\frac{1}{x}}\right) = \lim_{u \to 0^+}\left(\frac{3 u + 4}{u}\right)$$
=
$$\frac{0 \cdot 3 + 4}{0} = \infty$$
The final answer:
$$\lim_{x \to \infty}\left(4 x + 3\right) = \infty$$
$$\lim_{x \to \infty}\left(4 x + 3\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(4 x + 3\right)$$ =
$$\lim_{x \to \infty}\left(\frac{4 + \frac{3}{x}}{\frac{1}{x}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{4 + \frac{3}{x}}{\frac{1}{x}}\right) = \lim_{u \to 0^+}\left(\frac{3 u + 4}{u}\right)$$
=
$$\frac{0 \cdot 3 + 4}{0} = \infty$$
The final answer:
$$\lim_{x \to \infty}\left(4 x + 3\right) = \infty$$
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
$$\lim_{x \to \infty}\left(4 x + 3\right) = \infty$$
$$\lim_{x \to 0^-}\left(4 x + 3\right) = 3$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(4 x + 3\right) = 3$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(4 x + 3\right) = 7$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(4 x + 3\right) = 7$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(4 x + 3\right) = -\infty$$
More at x→-oo
$$\lim_{x \to 0^-}\left(4 x + 3\right) = 3$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(4 x + 3\right) = 3$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(4 x + 3\right) = 7$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(4 x + 3\right) = 7$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(4 x + 3\right) = -\infty$$
More at x→-oo
The graph