Mister Exam
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How to use it?
- Limit of the function:
-
Limit of (-5+x)^(1/(-6+x))
-
Limit of (-4+x^4+3*x^2)/(1+x^3-2*x)
-
Limit of -1+x^3*(x^4+2*x^3)
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Limit of x^3/(1+x^2-x)
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Identical expressions
- (- five +x)^(one /(- six +x))
- ( minus 5 plus x) to the power of (1 divide by ( minus 6 plus x))
- ( minus five plus x) to the power of (one divide by ( minus six plus x))
- (-5+x)(1/(-6+x))
- -5+x1/-6+x
- -5+x^1/-6+x
- (-5+x)^(1 divide by (-6+x))
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Similar expressions
- (5+x)^(1/(-6+x))
- (-5-x)^(1/(-6+x))
- (-5+x)^(1/(-6-x))
- (-5+x)^(1/(6+x))
Limit of the function (-5+x)^(1/(-6+x))
The solution
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[src]
1
------
-6 + x
lim (-5 + x)
x->6+
$$\lim_{x \to 6^+} \left(x - 5\right)^{\frac{1}{x - 6}}$$
Limit((-5 + x)^(1/(-6 + x)), x, 6)
Detail solution
Let's take the limit
$$\lim_{x \to 6^+} \left(x - 5\right)^{\frac{1}{x - 6}}$$
transform
do replacement
$$u = \frac{1}{x - 6}$$
then
$$\lim_{x \to 6^+} \left(1 + \frac{1}{\frac{1}{x - 6}}\right)^{\frac{1}{x - 6}}$$ =
=
$$\lim_{u \to 6^+} \left(1 + \frac{1}{u}\right)^{u}$$
=
$$\lim_{u \to 6^+} \left(1 + \frac{1}{u}\right)^{u}$$
=
$$\left(\left(\lim_{u \to 6^+} \left(1 + \frac{1}{u}\right)^{u}\right)\right)$$
The limit
$$\lim_{u \to 6^+} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to 6^+} \left(1 + \frac{1}{u}\right)^{u}\right)\right) = e$$
The final answer:
$$\lim_{x \to 6^+} \left(x - 5\right)^{\frac{1}{x - 6}} = e$$
$$\lim_{x \to 6^+} \left(x - 5\right)^{\frac{1}{x - 6}}$$
transform
do replacement
$$u = \frac{1}{x - 6}$$
then
$$\lim_{x \to 6^+} \left(1 + \frac{1}{\frac{1}{x - 6}}\right)^{\frac{1}{x - 6}}$$ =
=
$$\lim_{u \to 6^+} \left(1 + \frac{1}{u}\right)^{u}$$
=
$$\lim_{u \to 6^+} \left(1 + \frac{1}{u}\right)^{u}$$
=
$$\left(\left(\lim_{u \to 6^+} \left(1 + \frac{1}{u}\right)^{u}\right)\right)$$
The limit
$$\lim_{u \to 6^+} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to 6^+} \left(1 + \frac{1}{u}\right)^{u}\right)\right) = e$$
The final answer:
$$\lim_{x \to 6^+} \left(x - 5\right)^{\frac{1}{x - 6}} = e$$
Lopital's rule
There is no sense to apply Lopital's rule to this function since there is no indeterminateness of 0/0 or oo/oo type
The graph
One‐sided limits
[src]
1
------
-6 + x
lim (-5 + x)
x->6+
$$\lim_{x \to 6^+} \left(x - 5\right)^{\frac{1}{x - 6}}$$
E
$$e$$
= 2.71828182845905
1
------
-6 + x
lim (-5 + x)
x->6-
$$\lim_{x \to 6^-} \left(x - 5\right)^{\frac{1}{x - 6}}$$
E
$$e$$
= 2.71828182845905
= 2.71828182845905
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to 6^-} \left(x - 5\right)^{\frac{1}{x - 6}} = e$$
More at x→6 from the left
$$\lim_{x \to 6^+} \left(x - 5\right)^{\frac{1}{x - 6}} = e$$
$$\lim_{x \to \infty} \left(x - 5\right)^{\frac{1}{x - 6}} = 1$$
More at x→oo
$$\lim_{x \to 0^-} \left(x - 5\right)^{\frac{1}{x - 6}} = - \frac{\left(-5\right)^{\frac{5}{6}}}{5}$$
More at x→0 from the left
$$\lim_{x \to 0^+} \left(x - 5\right)^{\frac{1}{x - 6}} = - \frac{\left(-5\right)^{\frac{5}{6}}}{5}$$
More at x→0 from the right
$$\lim_{x \to 1^-} \left(x - 5\right)^{\frac{1}{x - 6}} = - \frac{\left(-1\right)^{\frac{4}{5}} \cdot 2^{\frac{3}{5}}}{2}$$
More at x→1 from the left
$$\lim_{x \to 1^+} \left(x - 5\right)^{\frac{1}{x - 6}} = - \frac{\left(-1\right)^{\frac{4}{5}} \cdot 2^{\frac{3}{5}}}{2}$$
More at x→1 from the right
$$\lim_{x \to -\infty} \left(x - 5\right)^{\frac{1}{x - 6}} = 1$$
More at x→-oo
More at x→6 from the left
$$\lim_{x \to 6^+} \left(x - 5\right)^{\frac{1}{x - 6}} = e$$
$$\lim_{x \to \infty} \left(x - 5\right)^{\frac{1}{x - 6}} = 1$$
More at x→oo
$$\lim_{x \to 0^-} \left(x - 5\right)^{\frac{1}{x - 6}} = - \frac{\left(-5\right)^{\frac{5}{6}}}{5}$$
More at x→0 from the left
$$\lim_{x \to 0^+} \left(x - 5\right)^{\frac{1}{x - 6}} = - \frac{\left(-5\right)^{\frac{5}{6}}}{5}$$
More at x→0 from the right
$$\lim_{x \to 1^-} \left(x - 5\right)^{\frac{1}{x - 6}} = - \frac{\left(-1\right)^{\frac{4}{5}} \cdot 2^{\frac{3}{5}}}{2}$$
More at x→1 from the left
$$\lim_{x \to 1^+} \left(x - 5\right)^{\frac{1}{x - 6}} = - \frac{\left(-1\right)^{\frac{4}{5}} \cdot 2^{\frac{3}{5}}}{2}$$
More at x→1 from the right
$$\lim_{x \to -\infty} \left(x - 5\right)^{\frac{1}{x - 6}} = 1$$
More at x→-oo
The graph