Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of x*2^x*3^(-x)
-
Limit of sin(5*x)/(4*x)
-
Limit of (-1+x^2)/(1-x)
- Limit of (a^x-a^b)/(x-b)
-
Identical expressions
- (- one +x^ two)/(one -x)
- ( minus 1 plus x squared ) divide by (1 minus x)
- ( minus one plus x to the power of two) divide by (one minus x)
- (-1+x2)/(1-x)
- -1+x2/1-x
- (-1+x²)/(1-x)
- (-1+x to the power of 2)/(1-x)
- -1+x^2/1-x
- (-1+x^2) divide by (1-x)
-
Similar expressions
- (1+x^2)/(1-x)
- (-1+x^2)/(1+x)
- (-1-x^2)/(1-x)
Limit of the function (-1+x^2)/(1-x)
The solution
You have entered
[src]
/ 2\
|-1 + x |
lim |-------|
x->1+\ 1 - x /
$$\lim_{x \to 1^+}\left(\frac{x^{2} - 1}{1 - x}\right)$$
Limit((-1 + x^2)/(1 - x), x, 1)
Detail solution
Let's take the limit
$$\lim_{x \to 1^+}\left(\frac{x^{2} - 1}{1 - x}\right)$$
transform
$$\lim_{x \to 1^+}\left(\frac{x^{2} - 1}{1 - x}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{\left(x - 1\right) \left(x + 1\right)}{1 - x}\right)$$
=
$$\lim_{x \to 1^+}\left(- x - 1\right) = $$
$$-1 - 1 = $$
The final answer:
$$\lim_{x \to 1^+}\left(\frac{x^{2} - 1}{1 - x}\right) = -2$$
$$\lim_{x \to 1^+}\left(\frac{x^{2} - 1}{1 - x}\right)$$
transform
$$\lim_{x \to 1^+}\left(\frac{x^{2} - 1}{1 - x}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{\left(x - 1\right) \left(x + 1\right)}{1 - x}\right)$$
=
$$\lim_{x \to 1^+}\left(- x - 1\right) = $$
$$-1 - 1 = $$
= -2
The final answer:
$$\lim_{x \to 1^+}\left(\frac{x^{2} - 1}{1 - x}\right) = -2$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to 1^+}\left(x^{2} - 1\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 1^+}\left(1 - x\right) = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 1^+}\left(\frac{x^{2} - 1}{1 - x}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{\frac{d}{d x} \left(x^{2} - 1\right)}{\frac{d}{d x} \left(1 - x\right)}\right)$$
=
$$\lim_{x \to 1^+}\left(- 2 x\right)$$
=
$$\lim_{x \to 1^+} -2$$
=
$$\lim_{x \to 1^+} -2$$
=
$$-2$$
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
$$\lim_{x \to 1^+}\left(x^{2} - 1\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 1^+}\left(1 - x\right) = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 1^+}\left(\frac{x^{2} - 1}{1 - x}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{\frac{d}{d x} \left(x^{2} - 1\right)}{\frac{d}{d x} \left(1 - x\right)}\right)$$
=
$$\lim_{x \to 1^+}\left(- 2 x\right)$$
=
$$\lim_{x \to 1^+} -2$$
=
$$\lim_{x \to 1^+} -2$$
=
$$-2$$
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]
/ 2\
|-1 + x |
lim |-------|
x->1+\ 1 - x /
$$\lim_{x \to 1^+}\left(\frac{x^{2} - 1}{1 - x}\right)$$
-2
$$-2$$
= -2.0
/ 2\
|-1 + x |
lim |-------|
x->1-\ 1 - x /
$$\lim_{x \to 1^-}\left(\frac{x^{2} - 1}{1 - x}\right)$$
-2
$$-2$$
= -2.0
= -2.0
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to 1^-}\left(\frac{x^{2} - 1}{1 - x}\right) = -2$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x^{2} - 1}{1 - x}\right) = -2$$
$$\lim_{x \to \infty}\left(\frac{x^{2} - 1}{1 - x}\right) = -\infty$$
More at x→oo
$$\lim_{x \to 0^-}\left(\frac{x^{2} - 1}{1 - x}\right) = -1$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x^{2} - 1}{1 - x}\right) = -1$$
More at x→0 from the right
$$\lim_{x \to -\infty}\left(\frac{x^{2} - 1}{1 - x}\right) = \infty$$
More at x→-oo
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x^{2} - 1}{1 - x}\right) = -2$$
$$\lim_{x \to \infty}\left(\frac{x^{2} - 1}{1 - x}\right) = -\infty$$
More at x→oo
$$\lim_{x \to 0^-}\left(\frac{x^{2} - 1}{1 - x}\right) = -1$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x^{2} - 1}{1 - x}\right) = -1$$
More at x→0 from the right
$$\lim_{x \to -\infty}\left(\frac{x^{2} - 1}{1 - x}\right) = \infty$$
More at x→-oo
The graph