Mister Exam
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- Limit of the function:
-
Limit of (x+log(1-x))/(1-x)
-
Limit of x*cos(x)
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Limit of x^2
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Limit of asin(1+x)^2/((1+x)*atan(1+x))
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Identical expressions
- (x+log(one -x))/(one -x)
- (x plus logarithm of (1 minus x)) divide by (1 minus x)
- (x plus logarithm of (one minus x)) divide by (one minus x)
- x+log1-x/1-x
- (x+log(1-x)) divide by (1-x)
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Similar expressions
- (x+log(1+x))/(1-x)
- (x+log(1-x))/(1+x)
- (x-log(1-x))/(1-x)
Limit of the function (x+log(1-x))/(1-x)
The solution
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/x + log(1 - x)\ lim |--------------| x->oo\ 1 - x /
$$\lim_{x \to \infty}\left(\frac{x + \log{\left(- x + 1 \right)}}{- x + 1}\right)$$
Limit((x + log(1 - x))/(1 - x), x, oo, dir='-')
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to \infty}\left(x + \log{\left(- x + 1 \right)}\right) = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(- x + 1\right) = -\infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{x + \log{\left(- x + 1 \right)}}{- x + 1}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(x + \log{\left(- x + 1 \right)}\right)}{\frac{d}{d x} \left(- x + 1\right)}\right)$$
=
$$\lim_{x \to \infty}\left(-1 + \frac{1}{- x + 1}\right)$$
=
$$\lim_{x \to \infty}\left(-1 + \frac{1}{- x + 1}\right)$$
=
$$-1$$
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
$$\lim_{x \to \infty}\left(x + \log{\left(- x + 1 \right)}\right) = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(- x + 1\right) = -\infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{x + \log{\left(- x + 1 \right)}}{- x + 1}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(x + \log{\left(- x + 1 \right)}\right)}{\frac{d}{d x} \left(- x + 1\right)}\right)$$
=
$$\lim_{x \to \infty}\left(-1 + \frac{1}{- x + 1}\right)$$
=
$$\lim_{x \to \infty}\left(-1 + \frac{1}{- x + 1}\right)$$
=
$$-1$$
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
$$\lim_{x \to \infty}\left(\frac{x + \log{\left(- x + 1 \right)}}{- x + 1}\right) = -1$$
$$\lim_{x \to 0^-}\left(\frac{x + \log{\left(- x + 1 \right)}}{- x + 1}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x + \log{\left(- x + 1 \right)}}{- x + 1}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x + \log{\left(- x + 1 \right)}}{- x + 1}\right) = -\infty$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x + \log{\left(- x + 1 \right)}}{- x + 1}\right) = \infty$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x + \log{\left(- x + 1 \right)}}{- x + 1}\right) = -1$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{x + \log{\left(- x + 1 \right)}}{- x + 1}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x + \log{\left(- x + 1 \right)}}{- x + 1}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x + \log{\left(- x + 1 \right)}}{- x + 1}\right) = -\infty$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x + \log{\left(- x + 1 \right)}}{- x + 1}\right) = \infty$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x + \log{\left(- x + 1 \right)}}{- x + 1}\right) = -1$$
More at x→-oo
The graph