limits, can they be done without using L'hopital's rule?

Question

can these limits be determined without usint L'Hopital's rule? Is there a way to compute them algebraically?

lim as x goes to 1 of (x-1)/(x-x^1/4)

lim as x goes to pi/2 of (5cosx)/(4x-2pi)

lim as x goes to 0 of (1+abs(x))^(1/x)

lim as x goes to 0 of x/(1-2^abs(x))

Answer 1

I like these sorts of problems, it a chance for creativity. Let's see:

#1: [x→1]lim (x-1)/(x - ∜x)

Make a change of variables: let y = ∜x. Then we have:

[y→1]lim (y⁴ - 1)/(y⁴ - y)

Doing some algebra:

[y→1]lim (y⁴ - y + y - 1)/(y⁴ - y)
[y→1]lim (y⁴ - y)/(y⁴ - y) + (y - 1)/(y⁴ - y)
[y→1]lim 1 + (y - 1)/(y⁴ - y)
[y→1]lim 1 + 1/((y⁴ - y)/(y - 1))

Using synthetic division on (y⁴ - y)/(y - 1):

1 | 1 0 0 -1 0
....... 1 1 1 0
-------------------
.... 1 1 1 0 0

[y→1]lim 1 + 1/(y³ + y² + y)

And this evaluates easily to:

1 + 1/(1 + 1 + 1)
4/3

#2: [x→π/2]lim (5 cos x)/(4x - 2π)

Again, make a change of variable, this time let y=x-π/2. Then this is:

[y→0]lim (5 cos (y+π/2))/(4y)

Using the angle addition formula for cosine:

[y→0]lim (5 (cos y cos (π/2) - sin y sin (π/2))/(4y)
[y→0]lim (-5 sin y)/(4y)
-5/4 [y→0]lim sin y/y
-5/4 (you know that [y→0]lim sin y/y = 1, since you computed it geometrically when deriving the fact that d(sin y)/dy = cos y)

#3: [x→0]lim (1 + |x|)^(1/x)

Fist we consider a simpler problem, [x→0]lim (1+ax)^(1/x). Since the exponential function is continuous, we can write this as:

e^([x→0]lim 1/x ln (1+ax))

Expand the natural logarithm using the Mercator series:

e^([x→0]lim 1/x [k=1, ∞]∑((-1)^(k-1)/k * (ax)^k))
e^([x→0]lim a [k=1, ∞]∑((-1)^(k-1)/k * (ax)^(k-1) ))
e^([x→0]lim a + a [k=2, ∞]∑((-1)^(k-1)/k * (ax)^(k-1) ))
e^a

So now we go back to the original problem, [x→0]lim (1 + |x|)^(1/x). We compute:

[x→0⁺]lim (1 + |x|)^(1/x) = [x→0⁺]lim (1 + x)^(1/x) = e
[x→0⁻]lim (1 + |x|)^(1/x) = [x→0⁺]lim (1 - x)^(1/x) = 1/e

Since the left and right-hand limits are not the same, [x→0]lim (1 + |x|)^(1/x) does not exist.

#4: [x→0]lim x/(1 - 2^|x|)

First we find the limit [x→0]lim x/(1 - e^(ax)). Using the power series for the exponential function:

[x→0]lim x/(1 - [k=0, ∞]∑((ax)^k/k!))
[x→0]lim x/(1 - 1 - [k=1, ∞]∑((ax)^k/k!))
[x→0]lim - x/[k=1, ∞]∑((ax)^k/k!)
[x→0]lim - 1/(a [k=1, ∞]∑((ax)^(k-1)/k!))
[x→0]lim - 1/(a + a [k=2, ∞]∑((ax)^(k-1)/k!))
-1/a

So we just evaluate the left and right-hand limits:

[x→0⁺]lim x/(1 - 2^|x|) = [x→0⁺]lim x/(1 - e^(x ln 2)) = -1/ln 2
[x→0⁻]lim x/(1 - 2^|x|) = [x→0⁺]lim x/(1 - e^(- x ln 2)) = 1/ln 2

Since the left and right-hand limits are not the same, this limit does not exist.

Answer 2

There are multiple ways to do pretty much any problem. That said, the second one is really easy by L'hopital's rule; I can't imagine another way would be easier yet. The same pretty much goes for the fourth one (just take the limit as x is positive, and wave your hands and say the values are identical when x is negative).

For the first one -- same story again.

For the third one, however, you're probably supposed to already know that the limit as x goes to infinity of (1 + 1/x)^x = e^x.