Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
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Limit of (1+3*x)^(5/x)
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Limit of (1+x)^(2/3)-(-1+x)^(2/3)
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Limit of 1/3+x/3
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Limit of e^(1+3*x)*(-1+x)
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Identical expressions
- (- one +x)/(- five +x^ two + four *x)
- ( minus 1 plus x) divide by ( minus 5 plus x squared plus 4 multiply by x)
- ( minus one plus x) divide by ( minus five plus x to the power of two plus four multiply by x)
- (-1+x)/(-5+x2+4*x)
- -1+x/-5+x2+4*x
- (-1+x)/(-5+x²+4*x)
- (-1+x)/(-5+x to the power of 2+4*x)
- (-1+x)/(-5+x^2+4x)
- (-1+x)/(-5+x2+4x)
- -1+x/-5+x2+4x
- -1+x/-5+x^2+4x
- (-1+x) divide by (-5+x^2+4*x)
-
Similar expressions
- (-1+x)/(-5-x^2+4*x)
- (-1+x)/(-5+x^2-4*x)
- (-1-x)/(-5+x^2+4*x)
- (-1+x)/(5+x^2+4*x)
- (1+x)/(-5+x^2+4*x)
Limit of the function (-1+x)/(-5+x^2+4*x)
The solution
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[src]
/ -1 + x \
lim |-------------|
x->1+| 2 |
\-5 + x + 4*x/
$$\lim_{x \to 1^+}\left(\frac{x - 1}{x^{2} + 4 x - 5}\right)$$
Limit((-1 + x)/(-5 + x^2 + 4*x), x, 1)
Detail solution
Let's take the limit
$$\lim_{x \to 1^+}\left(\frac{x - 1}{x^{2} + 4 x - 5}\right)$$
transform
$$\lim_{x \to 1^+}\left(\frac{x - 1}{x^{2} + 4 x - 5}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{x - 1}{\left(x - 1\right) \left(x + 5\right)}\right)$$
=
$$\lim_{x \to 1^+} \frac{1}{x + 5} = $$
$$\frac{1}{1 + 5} = $$
The final answer:
$$\lim_{x \to 1^+}\left(\frac{x - 1}{x^{2} + 4 x - 5}\right) = \frac{1}{6}$$
$$\lim_{x \to 1^+}\left(\frac{x - 1}{x^{2} + 4 x - 5}\right)$$
transform
$$\lim_{x \to 1^+}\left(\frac{x - 1}{x^{2} + 4 x - 5}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{x - 1}{\left(x - 1\right) \left(x + 5\right)}\right)$$
=
$$\lim_{x \to 1^+} \frac{1}{x + 5} = $$
$$\frac{1}{1 + 5} = $$
= 1/6
The final answer:
$$\lim_{x \to 1^+}\left(\frac{x - 1}{x^{2} + 4 x - 5}\right) = \frac{1}{6}$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to 1^+}\left(x - 1\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 1^+}\left(x^{2} + 4 x - 5\right) = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 1^+}\left(\frac{x - 1}{x^{2} + 4 x - 5}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{\frac{d}{d x} \left(x - 1\right)}{\frac{d}{d x} \left(x^{2} + 4 x - 5\right)}\right)$$
=
$$\lim_{x \to 1^+} \frac{1}{2 x + 4}$$
=
$$\lim_{x \to 1^+} \frac{1}{2 x + 4}$$
=
$$\frac{1}{6}$$
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
$$\lim_{x \to 1^+}\left(x - 1\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 1^+}\left(x^{2} + 4 x - 5\right) = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 1^+}\left(\frac{x - 1}{x^{2} + 4 x - 5}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{\frac{d}{d x} \left(x - 1\right)}{\frac{d}{d x} \left(x^{2} + 4 x - 5\right)}\right)$$
=
$$\lim_{x \to 1^+} \frac{1}{2 x + 4}$$
=
$$\lim_{x \to 1^+} \frac{1}{2 x + 4}$$
=
$$\frac{1}{6}$$
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
$$\lim_{x \to 1^-}\left(\frac{x - 1}{x^{2} + 4 x - 5}\right) = \frac{1}{6}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x - 1}{x^{2} + 4 x - 5}\right) = \frac{1}{6}$$
$$\lim_{x \to \infty}\left(\frac{x - 1}{x^{2} + 4 x - 5}\right) = 0$$
More at x→oo
$$\lim_{x \to 0^-}\left(\frac{x - 1}{x^{2} + 4 x - 5}\right) = \frac{1}{5}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x - 1}{x^{2} + 4 x - 5}\right) = \frac{1}{5}$$
More at x→0 from the right
$$\lim_{x \to -\infty}\left(\frac{x - 1}{x^{2} + 4 x - 5}\right) = 0$$
More at x→-oo
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x - 1}{x^{2} + 4 x - 5}\right) = \frac{1}{6}$$
$$\lim_{x \to \infty}\left(\frac{x - 1}{x^{2} + 4 x - 5}\right) = 0$$
More at x→oo
$$\lim_{x \to 0^-}\left(\frac{x - 1}{x^{2} + 4 x - 5}\right) = \frac{1}{5}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x - 1}{x^{2} + 4 x - 5}\right) = \frac{1}{5}$$
More at x→0 from the right
$$\lim_{x \to -\infty}\left(\frac{x - 1}{x^{2} + 4 x - 5}\right) = 0$$
More at x→-oo
One‐sided limits
[src]
/ -1 + x \
lim |-------------|
x->1+| 2 |
\-5 + x + 4*x/
$$\lim_{x \to 1^+}\left(\frac{x - 1}{x^{2} + 4 x - 5}\right)$$
1/6
$$\frac{1}{6}$$
= 0.166666666666667
/ -1 + x \
lim |-------------|
x->1-| 2 |
\-5 + x + 4*x/
$$\lim_{x \to 1^-}\left(\frac{x - 1}{x^{2} + 4 x - 5}\right)$$
1/6
$$\frac{1}{6}$$
= 0.166666666666667
= 0.166666666666667
The graph