Mister Exam
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How to use it?
- Limit of the function:
-
Limit of ((2+x)^2-(-2+x)^2)/(2+x)
-
Limit of -2/7+x^2/7
-
Limit of (1+x^2-2*x)/(1-4*x+3*x^2)
- Limit of (1+x^2+2*x)/(2+x^2+3*x)
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Identical expressions
- (- one +e^(five *x))/x
- ( minus 1 plus e to the power of (5 multiply by x)) divide by x
- ( minus one plus e to the power of (five multiply by x)) divide by x
- (-1+e(5*x))/x
- -1+e5*x/x
- (-1+e^(5x))/x
- (-1+e(5x))/x
- -1+e5x/x
- -1+e^5x/x
- (-1+e^(5*x)) divide by x
-
Similar expressions
- (-1-e^(5*x))/x
- (1+e^(5*x))/x
Limit of the function (-1+e^(5*x))/x
The solution
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[src]
/ 5*x\
|-1 + E |
lim |---------|
x->0+\ x /
$$\lim_{x \to 0^+}\left(\frac{e^{5 x} - 1}{x}\right)$$
Limit((-1 + E^(5*x))/x, x, 0)
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to 0^+}\left(e^{5 x} - 1\right) = 0$$
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.
$$\lim_{x \to 0^+}\left(\frac{e^{5 x} - 1}{x}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to 0^+}\left(\frac{e^{5 x} - 1}{x}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(e^{5 x} - 1\right)}{\frac{d}{d x} x}\right)$$
=
$$\lim_{x \to 0^+}\left(5 e^{5 x}\right)$$
=
$$\lim_{x \to 0^+} 5$$
=
$$\lim_{x \to 0^+} 5$$
=
$$5$$
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
$$\lim_{x \to 0^+}\left(e^{5 x} - 1\right) = 0$$
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.
$$\lim_{x \to 0^+}\left(\frac{e^{5 x} - 1}{x}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to 0^+}\left(\frac{e^{5 x} - 1}{x}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(e^{5 x} - 1\right)}{\frac{d}{d x} x}\right)$$
=
$$\lim_{x \to 0^+}\left(5 e^{5 x}\right)$$
=
$$\lim_{x \to 0^+} 5$$
=
$$\lim_{x \to 0^+} 5$$
=
$$5$$
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
$$\lim_{x \to 0^-}\left(\frac{e^{5 x} - 1}{x}\right) = 5$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{e^{5 x} - 1}{x}\right) = 5$$
$$\lim_{x \to \infty}\left(\frac{e^{5 x} - 1}{x}\right) = \infty$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{e^{5 x} - 1}{x}\right) = -1 + e^{5}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{e^{5 x} - 1}{x}\right) = -1 + e^{5}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{e^{5 x} - 1}{x}\right) = 0$$
More at x→-oo
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{e^{5 x} - 1}{x}\right) = 5$$
$$\lim_{x \to \infty}\left(\frac{e^{5 x} - 1}{x}\right) = \infty$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{e^{5 x} - 1}{x}\right) = -1 + e^{5}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{e^{5 x} - 1}{x}\right) = -1 + e^{5}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{e^{5 x} - 1}{x}\right) = 0$$
More at x→-oo
One‐sided limits
[src]
/ 5*x\
|-1 + E |
lim |---------|
x->0+\ x /
$$\lim_{x \to 0^+}\left(\frac{e^{5 x} - 1}{x}\right)$$
5
$$5$$
= 5.0
/ 5*x\
|-1 + E |
lim |---------|
x->0-\ x /
$$\lim_{x \to 0^-}\left(\frac{e^{5 x} - 1}{x}\right)$$
5
$$5$$
= 5.0
= 5.0
The graph