Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
-
Limit of 10+x^2+3*x^3+8*x
-
Limit of (-16+2^x)/(-1+5*sqrt(x)*(5-x))
-
Limit of (-4+x^2)/(-3+sqrt(1-4*x))
-
Limit of ((1+x)/(-2+x))^(3+x)
-
Identical expressions
- (four -x)/(- sixteen +x^ two)
- (4 minus x) divide by ( minus 16 plus x squared )
- (four minus x) divide by ( minus sixteen plus x to the power of two)
- (4-x)/(-16+x2)
- 4-x/-16+x2
- (4-x)/(-16+x²)
- (4-x)/(-16+x to the power of 2)
- 4-x/-16+x^2
- (4-x) divide by (-16+x^2)
-
Similar expressions
- (4+x)/(-16+x^2)
- (4-x)/(16+x^2)
- (4-x)/(-16-x^2)
Limit of the function (4-x)/(-16+x^2)
The solution
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[src]
/ 4 - x \
lim |--------|
x->4+| 2|
\-16 + x /
$$\lim_{x \to 4^+}\left(\frac{4 - x}{x^{2} - 16}\right)$$
Limit((4 - x)/(-16 + x^2), x, 4)
Detail solution
Let's take the limit
$$\lim_{x \to 4^+}\left(\frac{4 - x}{x^{2} - 16}\right)$$
transform
$$\lim_{x \to 4^+}\left(\frac{4 - x}{x^{2} - 16}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{4 - x}{\left(x - 4\right) \left(x + 4\right)}\right)$$
=
$$\lim_{x \to 4^+}\left(- \frac{1}{x + 4}\right) = $$
$$- \frac{1}{4 + 4} = $$
The final answer:
$$\lim_{x \to 4^+}\left(\frac{4 - x}{x^{2} - 16}\right) = - \frac{1}{8}$$
$$\lim_{x \to 4^+}\left(\frac{4 - x}{x^{2} - 16}\right)$$
transform
$$\lim_{x \to 4^+}\left(\frac{4 - x}{x^{2} - 16}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{4 - x}{\left(x - 4\right) \left(x + 4\right)}\right)$$
=
$$\lim_{x \to 4^+}\left(- \frac{1}{x + 4}\right) = $$
$$- \frac{1}{4 + 4} = $$
= -1/8
The final answer:
$$\lim_{x \to 4^+}\left(\frac{4 - x}{x^{2} - 16}\right) = - \frac{1}{8}$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to 4^+}\left(4 - x\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 4^+}\left(x^{2} - 16\right) = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 4^+}\left(\frac{4 - x}{x^{2} - 16}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{\frac{d}{d x} \left(4 - x\right)}{\frac{d}{d x} \left(x^{2} - 16\right)}\right)$$
=
$$\lim_{x \to 4^+}\left(- \frac{1}{2 x}\right)$$
=
$$\lim_{x \to 4^+} - \frac{1}{8}$$
=
$$\lim_{x \to 4^+} - \frac{1}{8}$$
=
$$- \frac{1}{8}$$
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
$$\lim_{x \to 4^+}\left(4 - x\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 4^+}\left(x^{2} - 16\right) = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 4^+}\left(\frac{4 - x}{x^{2} - 16}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{\frac{d}{d x} \left(4 - x\right)}{\frac{d}{d x} \left(x^{2} - 16\right)}\right)$$
=
$$\lim_{x \to 4^+}\left(- \frac{1}{2 x}\right)$$
=
$$\lim_{x \to 4^+} - \frac{1}{8}$$
=
$$\lim_{x \to 4^+} - \frac{1}{8}$$
=
$$- \frac{1}{8}$$
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
$$\lim_{x \to 4^-}\left(\frac{4 - x}{x^{2} - 16}\right) = - \frac{1}{8}$$
More at x→4 from the left
$$\lim_{x \to 4^+}\left(\frac{4 - x}{x^{2} - 16}\right) = - \frac{1}{8}$$
$$\lim_{x \to \infty}\left(\frac{4 - x}{x^{2} - 16}\right) = 0$$
More at x→oo
$$\lim_{x \to 0^-}\left(\frac{4 - x}{x^{2} - 16}\right) = - \frac{1}{4}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{4 - x}{x^{2} - 16}\right) = - \frac{1}{4}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{4 - x}{x^{2} - 16}\right) = - \frac{1}{5}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{4 - x}{x^{2} - 16}\right) = - \frac{1}{5}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{4 - x}{x^{2} - 16}\right) = 0$$
More at x→-oo
More at x→4 from the left
$$\lim_{x \to 4^+}\left(\frac{4 - x}{x^{2} - 16}\right) = - \frac{1}{8}$$
$$\lim_{x \to \infty}\left(\frac{4 - x}{x^{2} - 16}\right) = 0$$
More at x→oo
$$\lim_{x \to 0^-}\left(\frac{4 - x}{x^{2} - 16}\right) = - \frac{1}{4}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{4 - x}{x^{2} - 16}\right) = - \frac{1}{4}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{4 - x}{x^{2} - 16}\right) = - \frac{1}{5}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{4 - x}{x^{2} - 16}\right) = - \frac{1}{5}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{4 - x}{x^{2} - 16}\right) = 0$$
More at x→-oo
One‐sided limits
[src]
/ 4 - x \
lim |--------|
x->4+| 2|
\-16 + x /
$$\lim_{x \to 4^+}\left(\frac{4 - x}{x^{2} - 16}\right)$$
-1/8
$$- \frac{1}{8}$$
= -0.125
/ 4 - x \
lim |--------|
x->4-| 2|
\-16 + x /
$$\lim_{x \to 4^-}\left(\frac{4 - x}{x^{2} - 16}\right)$$
-1/8
$$- \frac{1}{8}$$
= -0.125
= -0.125
The graph