Mister Exam
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How to use it?
- Limit of the function:
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Limit of (-5+7*n)/(2+n)
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Limit of (-2+x)/(-4+x)
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Limit of (-16+x^2-6*x)/(-2+x+x^2)
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Limit of (2+x^2-5*x)/(5+x^2+3*x^3)
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Identical expressions
- ((one +x)^ two -sqrt(one + four *x))/(-e^(x^ two)+cos(x))
- ((1 plus x) squared minus square root of (1 plus 4 multiply by x)) divide by ( minus e to the power of (x squared ) plus co sinus of e of (x))
- ((one plus x) to the power of two minus square root of (one plus four multiply by x)) divide by ( minus e to the power of (x to the power of two) plus co sinus of e of (x))
- ((1+x)^2-√(1+4*x))/(-e^(x^2)+cos(x))
- ((1+x)2-sqrt(1+4*x))/(-e(x2)+cos(x))
- 1+x2-sqrt1+4*x/-ex2+cosx
- ((1+x)²-sqrt(1+4*x))/(-e^(x²)+cos(x))
- ((1+x) to the power of 2-sqrt(1+4*x))/(-e to the power of (x to the power of 2)+cos(x))
- ((1+x)^2-sqrt(1+4x))/(-e^(x^2)+cos(x))
- ((1+x)2-sqrt(1+4x))/(-e(x2)+cos(x))
- 1+x2-sqrt1+4x/-ex2+cosx
- 1+x^2-sqrt1+4x/-e^x^2+cosx
- ((1+x)^2-sqrt(1+4*x)) divide by (-e^(x^2)+cos(x))
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Similar expressions
- ((1+x)^2-sqrt(1+4*x))/(-e^(x^2)-cos(x))
- ((1-x)^2-sqrt(1+4*x))/(-e^(x^2)+cos(x))
- ((1+x)^2+sqrt(1+4*x))/(-e^(x^2)+cos(x))
- ((1+x)^2-sqrt(1-4*x))/(-e^(x^2)+cos(x))
- ((1+x)^2-sqrt(1+4*x))/(e^(x^2)+cos(x))
- ((1+x)^2-sqrt(1+4*x))/(-e^(x^2)+cosx)
Limit of the function ((1+x)^2-sqrt(1+4*x))/(-e^(x^2)+cos(x))
The solution
You have entered
[src]
/ 2 _________\
|(1 + x) - \/ 1 + 4*x |
lim |----------------------|
x->0+| / 2\ |
| \x / |
\ - E + cos(x) /
$$\lim_{x \to 0^+}\left(\frac{\left(x + 1\right)^{2} - \sqrt{4 x + 1}}{- e^{x^{2}} + \cos{\left(x \right)}}\right)$$
Limit(((1 + x)^2 - sqrt(1 + 4*x))/(-E^(x^2) + cos(x)), x, 0)
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to 0^+}\left(x^{2} + 2 x - \sqrt{4 x + 1} + 1\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+}\left(- e^{x^{2}} + \cos{\left(x \right)}\right) = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\left(x + 1\right)^{2} - \sqrt{4 x + 1}}{- e^{x^{2}} + \cos{\left(x \right)}}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to 0^+}\left(\frac{\left(x + 1\right)^{2} - \sqrt{4 x + 1}}{- e^{x^{2}} + \cos{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(x^{2} + 2 x - \sqrt{4 x + 1} + 1\right)}{\frac{d}{d x} \left(- e^{x^{2}} + \cos{\left(x \right)}\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2 x + 2 - \frac{2}{\sqrt{4 x + 1}}}{- 2 x e^{x^{2}} - \sin{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(2 x + 2 - \frac{2}{\sqrt{4 x + 1}}\right)}{\frac{d}{d x} \left(- 2 x e^{x^{2}} - \sin{\left(x \right)}\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2 + \frac{4}{4 x \sqrt{4 x + 1} + \sqrt{4 x + 1}}}{- 4 x^{2} e^{x^{2}} - 2 e^{x^{2}} - \cos{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2 + \frac{4}{4 x \sqrt{4 x + 1} + \sqrt{4 x + 1}}}{- 4 x^{2} e^{x^{2}} - 2 e^{x^{2}} - \cos{\left(x \right)}}\right)$$
=
$$-2$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 2 time(s)
0/0,
i.e. limit for the numerator is
$$\lim_{x \to 0^+}\left(x^{2} + 2 x - \sqrt{4 x + 1} + 1\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+}\left(- e^{x^{2}} + \cos{\left(x \right)}\right) = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\left(x + 1\right)^{2} - \sqrt{4 x + 1}}{- e^{x^{2}} + \cos{\left(x \right)}}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to 0^+}\left(\frac{\left(x + 1\right)^{2} - \sqrt{4 x + 1}}{- e^{x^{2}} + \cos{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(x^{2} + 2 x - \sqrt{4 x + 1} + 1\right)}{\frac{d}{d x} \left(- e^{x^{2}} + \cos{\left(x \right)}\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2 x + 2 - \frac{2}{\sqrt{4 x + 1}}}{- 2 x e^{x^{2}} - \sin{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(2 x + 2 - \frac{2}{\sqrt{4 x + 1}}\right)}{\frac{d}{d x} \left(- 2 x e^{x^{2}} - \sin{\left(x \right)}\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2 + \frac{4}{4 x \sqrt{4 x + 1} + \sqrt{4 x + 1}}}{- 4 x^{2} e^{x^{2}} - 2 e^{x^{2}} - \cos{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2 + \frac{4}{4 x \sqrt{4 x + 1} + \sqrt{4 x + 1}}}{- 4 x^{2} e^{x^{2}} - 2 e^{x^{2}} - \cos{\left(x \right)}}\right)$$
=
$$-2$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 2 time(s)
The graph
One‐sided limits
[src]
/ 2 _________\
|(1 + x) - \/ 1 + 4*x |
lim |----------------------|
x->0+| / 2\ |
| \x / |
\ - E + cos(x) /
$$\lim_{x \to 0^+}\left(\frac{\left(x + 1\right)^{2} - \sqrt{4 x + 1}}{- e^{x^{2}} + \cos{\left(x \right)}}\right)$$
-2
$$-2$$
= -2.0
/ 2 _________\
|(1 + x) - \/ 1 + 4*x |
lim |----------------------|
x->0-| / 2\ |
| \x / |
\ - E + cos(x) /
$$\lim_{x \to 0^-}\left(\frac{\left(x + 1\right)^{2} - \sqrt{4 x + 1}}{- e^{x^{2}} + \cos{\left(x \right)}}\right)$$
-2
$$-2$$
= -2.0
= -2.0
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to 0^-}\left(\frac{\left(x + 1\right)^{2} - \sqrt{4 x + 1}}{- e^{x^{2}} + \cos{\left(x \right)}}\right) = -2$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\left(x + 1\right)^{2} - \sqrt{4 x + 1}}{- e^{x^{2}} + \cos{\left(x \right)}}\right) = -2$$
$$\lim_{x \to \infty}\left(\frac{\left(x + 1\right)^{2} - \sqrt{4 x + 1}}{- e^{x^{2}} + \cos{\left(x \right)}}\right) = 0$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{\left(x + 1\right)^{2} - \sqrt{4 x + 1}}{- e^{x^{2}} + \cos{\left(x \right)}}\right) = \frac{-4 + \sqrt{5}}{e - \cos{\left(1 \right)}}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\left(x + 1\right)^{2} - \sqrt{4 x + 1}}{- e^{x^{2}} + \cos{\left(x \right)}}\right) = \frac{-4 + \sqrt{5}}{e - \cos{\left(1 \right)}}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\left(x + 1\right)^{2} - \sqrt{4 x + 1}}{- e^{x^{2}} + \cos{\left(x \right)}}\right) = 0$$
More at x→-oo
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\left(x + 1\right)^{2} - \sqrt{4 x + 1}}{- e^{x^{2}} + \cos{\left(x \right)}}\right) = -2$$
$$\lim_{x \to \infty}\left(\frac{\left(x + 1\right)^{2} - \sqrt{4 x + 1}}{- e^{x^{2}} + \cos{\left(x \right)}}\right) = 0$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{\left(x + 1\right)^{2} - \sqrt{4 x + 1}}{- e^{x^{2}} + \cos{\left(x \right)}}\right) = \frac{-4 + \sqrt{5}}{e - \cos{\left(1 \right)}}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\left(x + 1\right)^{2} - \sqrt{4 x + 1}}{- e^{x^{2}} + \cos{\left(x \right)}}\right) = \frac{-4 + \sqrt{5}}{e - \cos{\left(1 \right)}}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\left(x + 1\right)^{2} - \sqrt{4 x + 1}}{- e^{x^{2}} + \cos{\left(x \right)}}\right) = 0$$
More at x→-oo
The graph