Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
-
Limit of (-log(1+x^2)+sin(x))/(2-e^(3*x)-sqrt(1+x))
-
Limit of sin(2*x)/log(1+x)
-
Limit of sin(21*x)/(3*x)
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Limit of (-2*sin(x)+sin(2*x))/x
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Identical expressions
- sin(two *x)/log(one +x)
- sinus of (2 multiply by x) divide by logarithm of (1 plus x)
- sinus of (two multiply by x) divide by logarithm of (one plus x)
- sin(2x)/log(1+x)
- sin2x/log1+x
- sin(2*x) divide by log(1+x)
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Similar expressions
- sin(2*x)/log(1-x)
- x*sin(2*x)/log(1+x^2)
Limit of the function sin(2*x)/log(1+x)
The solution
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[src]
/ sin(2*x) \ lim |----------| x->0+\log(1 + x)/
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\log{\left(x + 1 \right)}}\right)$$
Limit(sin(2*x)/log(1 + x), x, 0)
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to 0^+} \sin{\left(2 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \log{\left(x + 1 \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\log{\left(x + 1 \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(2 x \right)}}{\frac{d}{d x} \log{\left(x + 1 \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(2 \left(x + 1\right) \cos{\left(2 x \right)}\right)$$
=
$$\lim_{x \to 0^+}\left(2 x + 2\right)$$
=
$$\lim_{x \to 0^+}\left(2 x + 2\right)$$
=
$$2$$
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
$$\lim_{x \to 0^+} \sin{\left(2 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \log{\left(x + 1 \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\log{\left(x + 1 \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(2 x \right)}}{\frac{d}{d x} \log{\left(x + 1 \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(2 \left(x + 1\right) \cos{\left(2 x \right)}\right)$$
=
$$\lim_{x \to 0^+}\left(2 x + 2\right)$$
=
$$\lim_{x \to 0^+}\left(2 x + 2\right)$$
=
$$2$$
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
One‐sided limits
[src]
/ sin(2*x) \ lim |----------| x->0+\log(1 + x)/
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\log{\left(x + 1 \right)}}\right)$$
2
$$2$$
= 2.0
/ sin(2*x) \ lim |----------| x->0-\log(1 + x)/
$$\lim_{x \to 0^-}\left(\frac{\sin{\left(2 x \right)}}{\log{\left(x + 1 \right)}}\right)$$
2
$$2$$
= 2.0
= 2.0
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to 0^-}\left(\frac{\sin{\left(2 x \right)}}{\log{\left(x + 1 \right)}}\right) = 2$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\log{\left(x + 1 \right)}}\right) = 2$$
$$\lim_{x \to \infty}\left(\frac{\sin{\left(2 x \right)}}{\log{\left(x + 1 \right)}}\right) = 0$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{\sin{\left(2 x \right)}}{\log{\left(x + 1 \right)}}\right) = \frac{\sin{\left(2 \right)}}{\log{\left(2 \right)}}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\sin{\left(2 x \right)}}{\log{\left(x + 1 \right)}}\right) = \frac{\sin{\left(2 \right)}}{\log{\left(2 \right)}}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\sin{\left(2 x \right)}}{\log{\left(x + 1 \right)}}\right) = 0$$
More at x→-oo
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\log{\left(x + 1 \right)}}\right) = 2$$
$$\lim_{x \to \infty}\left(\frac{\sin{\left(2 x \right)}}{\log{\left(x + 1 \right)}}\right) = 0$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{\sin{\left(2 x \right)}}{\log{\left(x + 1 \right)}}\right) = \frac{\sin{\left(2 \right)}}{\log{\left(2 \right)}}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\sin{\left(2 x \right)}}{\log{\left(x + 1 \right)}}\right) = \frac{\sin{\left(2 \right)}}{\log{\left(2 \right)}}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\sin{\left(2 x \right)}}{\log{\left(x + 1 \right)}}\right) = 0$$
More at x→-oo
The graph