Simple limit (lim) problem ?

Question

I am not sure the answer and would like to check with someone about this question:

http://img139.imageshack.us/img139/3452/limitfu3.jpg

Thanks!

Answer 1

i got 2

Answer 2

Here's my answer

http://files.myopera.com/sthnew/files/li...

Answer 3

an easier way to find this limit is to take the series expansion for the numerator...for small x, exp(2x)=1+2x+...

so the limit becomes (1+2x-1)/x and approaches 2 as x approaches zero

Answer 4

The answer is 2. This can be obtained since the limit is an indeterminate form. Hence, l'Hopital rule can be applied:

lim f(x) / g(x) = lim f'(x)/g'(x)

x->k                x->k

Answer 5

It's hard to evaluate because it looks like 0/0.

The limit is actually 2.

L'Hopital's rule states that for functions f(x) and g(x),

lim f(x) / g(x) = lim f'(x)/g'(x)

x->k

Differentiate the numerator and denominator and evaluate again:

d/dx (e^(2x)-1) = 2e^x

d/dx x = 1

2e^x/1

For x=0

2e^0/1 = 2

Answer 6

Explain why we cannot easily evaluate the limit below.  Can you suggest an approach for evaluating the limit?

lim [ e^(2x) - 1 ] / [ x ]

x -> 0

We cannot easily evaluate the limit (at least, pre-Calculus II when we can apply L'Hospital's rule) is because we are not able to cancel the x in the denominator with elementary algebraic manipulation.

However, we know that y = e^x has the property that, at x = 0, the slope of the tangent is equal to 1.  This means since the definition of the derivative is

f'(x) = lim [ f(x + h) - f(x) ] / h

. . . . . h -> 0

It means that for f(x) = e^x, and x = 0,

1 = f'(0) = lim [ e^(0 + h) - e^0 ] / h

. . . . . . . . h -> 0

1 = lim [ e^h - 1 ] / h

. . . h -> 0

Back to the original question:

lim [ (e^x)^2 - 1 ] / [ x ]

x -> 0

We have a difference of squares in the numerator, which we can factor.

lim [ (e^x - 1) (e^x + 1) ] / x

x -> 0

Which we can turn into the limit of a product.

lim ( [ (e^x - 1)/x ] [ e^x + 1 ] )

x -> 0

lim [ (e^x - 1)/x ] lim  [ e^x + 1 ]

x -> 0 . . . . . . . . x -> 0

But by our above knowledge of what occurs at x = 0 for the slope of e^x,

lim [ (e^x - 1)/x ] = 1

x -> 0

So we get

(1) lim [ e^x + 1]

. . . x -> 0

Which we can now directly evaluate.

[ e^0 + 1 ]

1 + 1

2

Answer 7

You cannot easily evaluate that limit because when you plug in zero, you get 0 in the numerator and zero in the denominator. Solutions of 0/0 are indeterminate.

Your approach depends on how much you have learned about limits.

The best approach would be l'Hôpital's rule, which can be attempted when you get a fraction of 0/0 or ±∞/±∞.

The rule says that if a fraction takes one of these forms, then:

lim g(x)/h(x)

is equal to the first of the following to give a result that is not indeterminate:

lim g'(x)/h'(x)

lim g''(x)/h''(x)

lim g'''(x)/h'''(x)

That means that if you get an indeterminate form, try getting the derivative of the numerator and denominator to find the solution. This works because you are interested in the relative rate of change of the numerator and denominator at that point.

This method works because infinity and zero "wipe out" other factors when they are present; therefore, it becomes useful to know how fast the numerator is approaching those values (like if the numerator is approaching infinity twice as fast as the denominator). The rate that it approaches them is given by the derivative.

----

So take the derivatives:

[lim x → 0] [ e^(2·x) - 1 ] / x

= [lim x → 0] d/dx [ e^(2·x) - 1 ] / d/dx ( x )

= [lim x → 0] [ 2·e^(2·x) - 0 ] / ( 1 )

= [lim x → 0] 2·e^(2·x)

= 2·e^(2·0)

= 2·eº

= 2

——————————————————————————————————————

Another way of evaluating the limit requires an elaborate manipulation of e^(2·x) involving one of its definitions.

e is defined by:

e = [lim n → ∞] (1 + 1/n)ⁿ

If n becomes 1/h, this changes to an identical form:

e = [lim h → 0] (1 + h)^(1/h)

Which means that, for values of h small, there is an improving approximation of e given by:

e ≈ (1 + Δh)^(1/Δh)

Which can be rewritten:

e^(Δh) ≈ (1 + Δh)

----

If you rewrite the given limit:

[lim x → 0] [ e^(2·x) - 1 ] / x

= [lim x → 0] [ e^(x)² - 1 ] / x

you can then replace e^(x) with (1 + x) as this holds for small values of x (and is equal to the limit as x approaches zero).

= [lim x → 0] [ (1 + x)² - 1 ] / x

= [lim x → 0] [ (1 + 2·x + x²) - 1 ] / x

= [lim x → 0] [ 2·x + x² ] / x

= [lim x → 0] [ 2 + x ]·x / x

= [lim x → 0] ( 2 + x )

= 2 + 0

= 2