Mister Exam
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How to use it?
- Limit of the function:
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Limit of -7/5+2*x
-
Limit of (8-7*x+4*x^2)/(-5-4*x+6*x^2)
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Limit of ((-1+2*x)/(2*x))^x
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Limit of x*cot(6*x)
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Identical expressions
- - seven / five + two *x
- minus 7 divide by 5 plus 2 multiply by x
- minus seven divide by five plus two multiply by x
- -7/5+2x
- -7 divide by 5+2*x
-
Similar expressions
- 7/5+2*x
- -7/5-2*x
Limit of the function -7/5+2*x
The solution
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lim (-7/5 + 2*x) x->oo
$$\lim_{x \to \infty}\left(2 x - \frac{7}{5}\right)$$
Limit(-7/5 + 2*x, x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(2 x - \frac{7}{5}\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(2 x - \frac{7}{5}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{2 - \frac{7}{5 x}}{\frac{1}{x}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{2 - \frac{7}{5 x}}{\frac{1}{x}}\right) = \lim_{u \to 0^+}\left(\frac{2 - \frac{7 u}{5}}{u}\right)$$
=
$$\frac{2 - 0}{0} = \infty$$
The final answer:
$$\lim_{x \to \infty}\left(2 x - \frac{7}{5}\right) = \infty$$
$$\lim_{x \to \infty}\left(2 x - \frac{7}{5}\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(2 x - \frac{7}{5}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{2 - \frac{7}{5 x}}{\frac{1}{x}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{2 - \frac{7}{5 x}}{\frac{1}{x}}\right) = \lim_{u \to 0^+}\left(\frac{2 - \frac{7 u}{5}}{u}\right)$$
=
$$\frac{2 - 0}{0} = \infty$$
The final answer:
$$\lim_{x \to \infty}\left(2 x - \frac{7}{5}\right) = \infty$$
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
$$\lim_{x \to \infty}\left(2 x - \frac{7}{5}\right) = \infty$$
$$\lim_{x \to 0^-}\left(2 x - \frac{7}{5}\right) = - \frac{7}{5}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(2 x - \frac{7}{5}\right) = - \frac{7}{5}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(2 x - \frac{7}{5}\right) = \frac{3}{5}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(2 x - \frac{7}{5}\right) = \frac{3}{5}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(2 x - \frac{7}{5}\right) = -\infty$$
More at x→-oo
$$\lim_{x \to 0^-}\left(2 x - \frac{7}{5}\right) = - \frac{7}{5}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(2 x - \frac{7}{5}\right) = - \frac{7}{5}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(2 x - \frac{7}{5}\right) = \frac{3}{5}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(2 x - \frac{7}{5}\right) = \frac{3}{5}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(2 x - \frac{7}{5}\right) = -\infty$$
More at x→-oo
The graph