Mister Exam
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- Limit of the function:
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Limit of (2*(-1)^x+8*x^6)/(32-6*x^5+5*x^6)
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Limit of -3-x+11*x^2/2
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Limit of (-5-24*x+5*x^2)/(-5+x)
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Limit of (2+4*x)/(1+x)
-
Identical expressions
- (two + four *x)/(one +x)
- (2 plus 4 multiply by x) divide by (1 plus x)
- (two plus four multiply by x) divide by (one plus x)
- (2+4x)/(1+x)
- 2+4x/1+x
- (2+4*x) divide by (1+x)
-
Similar expressions
- (2-4*x)/(1+x)
- (3+x^2+4*x)/(1+x^2+2*x)
- (2+4*x)/(1-x)
- (3+x^2+4*x)/(1+x)
Limit of the function (2+4*x)/(1+x)
The solution
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/2 + 4*x\ lim |-------| x->oo\ 1 + x /
$$\lim_{x \to \infty}\left(\frac{4 x + 2}{x + 1}\right)$$
Limit((2 + 4*x)/(1 + x), x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(\frac{4 x + 2}{x + 1}\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(\frac{4 x + 2}{x + 1}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{4 + \frac{2}{x}}{1 + \frac{1}{x}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{4 + \frac{2}{x}}{1 + \frac{1}{x}}\right) = \lim_{u \to 0^+}\left(\frac{2 u + 4}{u + 1}\right)$$
=
$$\frac{0 \cdot 2 + 4}{1} = 4$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{4 x + 2}{x + 1}\right) = 4$$
$$\lim_{x \to \infty}\left(\frac{4 x + 2}{x + 1}\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(\frac{4 x + 2}{x + 1}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{4 + \frac{2}{x}}{1 + \frac{1}{x}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{4 + \frac{2}{x}}{1 + \frac{1}{x}}\right) = \lim_{u \to 0^+}\left(\frac{2 u + 4}{u + 1}\right)$$
=
$$\frac{0 \cdot 2 + 4}{1} = 4$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{4 x + 2}{x + 1}\right) = 4$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to \infty}\left(2 x + 1\right) = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(\frac{x}{2} + \frac{1}{2}\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{4 x + 2}{x + 1}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to \infty}\left(\frac{2 \left(2 x + 1\right)}{x + 1}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(2 x + 1\right)}{\frac{d}{d x} \left(\frac{x}{2} + \frac{1}{2}\right)}\right)$$
=
$$\lim_{x \to \infty} 4$$
=
$$\lim_{x \to \infty} 4$$
=
$$4$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)
oo/oo,
i.e. limit for the numerator is
$$\lim_{x \to \infty}\left(2 x + 1\right) = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(\frac{x}{2} + \frac{1}{2}\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{4 x + 2}{x + 1}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to \infty}\left(\frac{2 \left(2 x + 1\right)}{x + 1}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(2 x + 1\right)}{\frac{d}{d x} \left(\frac{x}{2} + \frac{1}{2}\right)}\right)$$
=
$$\lim_{x \to \infty} 4$$
=
$$\lim_{x \to \infty} 4$$
=
$$4$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)
The graph
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to \infty}\left(\frac{4 x + 2}{x + 1}\right) = 4$$
$$\lim_{x \to 0^-}\left(\frac{4 x + 2}{x + 1}\right) = 2$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{4 x + 2}{x + 1}\right) = 2$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{4 x + 2}{x + 1}\right) = 3$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{4 x + 2}{x + 1}\right) = 3$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{4 x + 2}{x + 1}\right) = 4$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{4 x + 2}{x + 1}\right) = 2$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{4 x + 2}{x + 1}\right) = 2$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{4 x + 2}{x + 1}\right) = 3$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{4 x + 2}{x + 1}\right) = 3$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{4 x + 2}{x + 1}\right) = 4$$
More at x→-oo
The graph