Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
-
Limit of ((-4+3*x)/(2+3*x))^(1+x)/3
-
Limit of (-16+2^x)/(-1+5*sqrt(x)*(5-x))
-
Limit of (-14+x^2-5*x)/(-6+x+2*x^2)
-
Limit of (3+x^2+4*x)/(1+x^3)
- Integral of d{x}:
- sqrt(4+x^2)/x
-
Identical expressions
- sqrt(four +x^ two)/x
- square root of (4 plus x squared ) divide by x
- square root of (four plus x to the power of two) divide by x
- √(4+x^2)/x
- sqrt(4+x2)/x
- sqrt4+x2/x
- sqrt(4+x²)/x
- sqrt(4+x to the power of 2)/x
- sqrt4+x^2/x
- sqrt(4+x^2) divide by x
-
Similar expressions
- sqrt(4-x^2)/x
Limit of the function sqrt(4+x^2)/x
The solution
You have entered
[src]
/ ________\
| / 2 |
|\/ 4 + x |
lim |-----------|
x->oo\ x /
$$\lim_{x \to \infty}\left(\frac{\sqrt{x^{2} + 4}}{x}\right)$$
Limit(sqrt(4 + x^2)/x, x, oo, dir='-')
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to \infty} \sqrt{x^{2} + 4} = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty} x = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{\sqrt{x^{2} + 4}}{x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \sqrt{x^{2} + 4}}{\frac{d}{d x} x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{x}{\sqrt{x^{2} + 4}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{x}{\sqrt{x^{2} + 4}}\right)$$
=
$$1$$
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)
oo/oo,
i.e. limit for the numerator is
$$\lim_{x \to \infty} \sqrt{x^{2} + 4} = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty} x = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{\sqrt{x^{2} + 4}}{x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \sqrt{x^{2} + 4}}{\frac{d}{d x} x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{x}{\sqrt{x^{2} + 4}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{x}{\sqrt{x^{2} + 4}}\right)$$
=
$$1$$
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
$$\lim_{x \to \infty}\left(\frac{\sqrt{x^{2} + 4}}{x}\right) = 1$$
$$\lim_{x \to 0^-}\left(\frac{\sqrt{x^{2} + 4}}{x}\right) = -\infty$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\sqrt{x^{2} + 4}}{x}\right) = \infty$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{\sqrt{x^{2} + 4}}{x}\right) = \sqrt{5}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\sqrt{x^{2} + 4}}{x}\right) = \sqrt{5}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\sqrt{x^{2} + 4}}{x}\right) = -1$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{\sqrt{x^{2} + 4}}{x}\right) = -\infty$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\sqrt{x^{2} + 4}}{x}\right) = \infty$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{\sqrt{x^{2} + 4}}{x}\right) = \sqrt{5}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\sqrt{x^{2} + 4}}{x}\right) = \sqrt{5}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\sqrt{x^{2} + 4}}{x}\right) = -1$$
More at x→-oo
The graph