Mister Exam
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How to use it?
- Limit of the function:
-
Limit of (3+x^2-4*x)/(-9+x^2)
-
Limit of (x^2-3*x)/(-8+x^2)
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Limit of (1+5*x)*(-1+5*x)
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Limit of (-x+tan(x))/(x+2*sin(x))
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Identical expressions
- (-log(two)+log(two +x))/x
- ( minus logarithm of (2) plus logarithm of (2 plus x)) divide by x
- ( minus logarithm of (two) plus logarithm of (two plus x)) divide by x
- -log2+log2+x/x
- (-log(2)+log(2+x)) divide by x
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Similar expressions
- (-log(2)+log(2-x))/x
- (log(2)+log(2+x))/x
- (-log(2)-log(2+x))/x
Limit of the function (-log(2)+log(2+x))/x
The solution
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[src]
/-log(2) + log(2 + x)\ lim |--------------------| x->0+\ x /
$$\lim_{x \to 0^+}\left(\frac{\log{\left(x + 2 \right)} - \log{\left(2 \right)}}{x}\right)$$
Limit((-log(2) + log(2 + x))/x, x, 0)
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to 0^+}\left(\log{\left(x + 2 \right)} - \log{\left(2 \right)}\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} x = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\log{\left(x + 2 \right)} - \log{\left(2 \right)}}{x}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(\log{\left(x + 2 \right)} - \log{\left(2 \right)}\right)}{\frac{d}{d x} x}\right)$$
=
$$\lim_{x \to 0^+} \frac{1}{x + 2}$$
=
$$\lim_{x \to 0^+} \frac{1}{x + 2}$$
=
$$\frac{1}{2}$$
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)
0/0,
i.e. limit for the numerator is
$$\lim_{x \to 0^+}\left(\log{\left(x + 2 \right)} - \log{\left(2 \right)}\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} x = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\log{\left(x + 2 \right)} - \log{\left(2 \right)}}{x}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(\log{\left(x + 2 \right)} - \log{\left(2 \right)}\right)}{\frac{d}{d x} x}\right)$$
=
$$\lim_{x \to 0^+} \frac{1}{x + 2}$$
=
$$\lim_{x \to 0^+} \frac{1}{x + 2}$$
=
$$\frac{1}{2}$$
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
$$\lim_{x \to 0^-}\left(\frac{\log{\left(x + 2 \right)} - \log{\left(2 \right)}}{x}\right) = \frac{1}{2}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\log{\left(x + 2 \right)} - \log{\left(2 \right)}}{x}\right) = \frac{1}{2}$$
$$\lim_{x \to \infty}\left(\frac{\log{\left(x + 2 \right)} - \log{\left(2 \right)}}{x}\right) = 0$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{\log{\left(x + 2 \right)} - \log{\left(2 \right)}}{x}\right) = - \log{\left(2 \right)} + \log{\left(3 \right)}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\log{\left(x + 2 \right)} - \log{\left(2 \right)}}{x}\right) = - \log{\left(2 \right)} + \log{\left(3 \right)}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\log{\left(x + 2 \right)} - \log{\left(2 \right)}}{x}\right) = 0$$
More at x→-oo
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\log{\left(x + 2 \right)} - \log{\left(2 \right)}}{x}\right) = \frac{1}{2}$$
$$\lim_{x \to \infty}\left(\frac{\log{\left(x + 2 \right)} - \log{\left(2 \right)}}{x}\right) = 0$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{\log{\left(x + 2 \right)} - \log{\left(2 \right)}}{x}\right) = - \log{\left(2 \right)} + \log{\left(3 \right)}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\log{\left(x + 2 \right)} - \log{\left(2 \right)}}{x}\right) = - \log{\left(2 \right)} + \log{\left(3 \right)}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\log{\left(x + 2 \right)} - \log{\left(2 \right)}}{x}\right) = 0$$
More at x→-oo
One‐sided limits
[src]
/-log(2) + log(2 + x)\ lim |--------------------| x->0+\ x /
$$\lim_{x \to 0^+}\left(\frac{\log{\left(x + 2 \right)} - \log{\left(2 \right)}}{x}\right)$$
1/2
$$\frac{1}{2}$$
= 0.5
/-log(2) + log(2 + x)\ lim |--------------------| x->0-\ x /
$$\lim_{x \to 0^-}\left(\frac{\log{\left(x + 2 \right)} - \log{\left(2 \right)}}{x}\right)$$
1/2
$$\frac{1}{2}$$
= 0.5
= 0.5
The graph