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3 years ago

Lim x-->0 ((sqrt(ax+b) -2) / x) =1 show steps....

1 answer:

4 0

Lim x→0 (√(ax+b)-2)/x=1

You want to know the value of "a" and "b"

lim x→0 (√(ax+b)-2)/x=(√(0+b)-2)/0=(√b -2)/0;

Then if (√b -2)/0=1; the numerator must be "0"
(√b-2)=0
√b=2
(√b)²=2²
b=4

It is necessary the numerator must be "0", if the denominator is "0" and the result is equal a number.

Therefore:
lim (√(ax+4)-2)/x=1
x⇒0

I imagine you know Taylor Series.
√(ax+4)=(4(1+ax/4))¹/²=2(1+ax/4)¹/²
Remember:
(1/2)
(1+x)ᵃ=<span>Σ ( a ) x^a
</span>
In our case:
(1/2) (1/2) (1/2)
(1+ax/4)¹/²=( 0) (ax/4)⁰+( 1 ) (ax/4)¹+( 2) (ax/4)²+...
=1 +(1/2) ax/4 + -1/8 (ax/4)²+...
=1+ax/8-a²x²/128+...

Therefore:
lim (√(ax+4)-2)/x=lim [2(1+ax/8-a²x²/128+...)-2]/x=
x⇒0 x⇒0

lim [(2+ax/4-a²x²/64+...)-2]/x=
x⇒0

lim (ax/4-a²x²/64+...)/x=
x⇒0

lim x(a/4-a²x/64+...)/x=
x⇒0

lim (a/4-a²x/64+...)=(a/4-0-0-0-...)=4/a
x⇒0

Because:

lim (√(ax+4)-2)/x=1
x⇒0

Then:
4/a=1 ⇒ a=4

Answer: a=4; b=4

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Determine whether each of the following functions is a solution of laplace's equation uxx uyy = 0.

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Both functions are the solution to the given Laplace solution.

Given Laplace's equation: $u_{x x}+u_{y y}=0$

We must determine whether a given function is the solution to a given Laplace equation.
If a function is a solution to a given Laplace's equation, it satisfies the solution.

(1) $u=e^{-x} \cos y-e^{-y} \cos x$

Differentiate with respect to x as follows:

$u_x=-e^{-x} \cos y+e^{-y} \sin x\\u_{x x}=e^{-x} \cos y+e^{-y} \cos x$

Differentiate with respect to y as follows:

$u_{x x}=e^{-x} \cos y+e^{-y} \cos x\\u_{y y}=-e^{-x} \cos y-e^{-y} \cos x$

Supplement the values in the given Laplace equation.

$e^{-x} \cos y+e^{-y} \cos x-e^{-x} \cos y-e^{-y} \cos x=0$

The given function in this case is the solution to the given Laplace equation.

(2) $u=\sin x \cosh y+\cos x \sinh y$

Differentiate with respect to x as follows:

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Differentiate with respect to y as follows:

$u_y=\sin x \sinh y+\cos x \cosh y\\u_{y y}=\sin x \cosh y+\cos x \sinh y$

Substitute the values to obtain:

$-\sin x \cosh y-\cos x \sinh y+\sin x \cosh y+\cos x \sinh y=0$
The given function in this case is the solution to the given Laplace equation.

Therefore, both functions are the solution to the given Laplace solution.

Know more about Laplace's equation here:

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The correct question is given below:
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