L'Hopital Rule

1.) What is the limit of ( e^x - (1+x) ) / x^n as x approaches 0 from the right?

Using the L'Hopital rule, I'm a little confused with regard to the denominator.

2.) What is the limit of (squareroot(25-x^2)) / (x^2 - 1) when x approaches 5 from the left?

Thank you in advance!

Hi!
Fir the first one you need to apple L'Hospital recursively until you rid the 'x' term in the denominator

(e^x - (1+x)) / xⁿ
so differentiate to get
(e^x - 1 ) / n * x^n-1
Repeat
(e^x) / n(n-1) * x^n-2

So now you know that if you keep differentiating, the numerator remains unaffected. However the denominator will reduce to something like
(n)(n-1)(n-2)......1 (= say N)

Now put the limiting value of x=0
you get
1/ N


Regarding the second one, i think the left hand limit will come out to be 0. Since the domain of the function is (-inf,5] , the LHL exists as x->5-

Hope it helps !

Quote:

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Originally Posted by TWN

1.) What is the limit of ( e^x - (1+x) ) / x^n as x approaches 0 from the right?

Using the L'Hopital rule, I'm a little confused with regard to the denominator.

2.) What is the limit of (squareroot(25-x^2)) / (x^2 - 1) when x approaches 5 from the left?

Thank you in advance!

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Surely you know that the derivative of x^n is nx^(n-1)...

Quote:

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Originally Posted by pratique21

Hi!
Fir the first one you need to apple L'Hospital recursively until you rid the 'x' term in the denominator

(e^x - (1+x)) / xⁿ
so differentiate to get
(e^x - 1 ) / n * x^n-1
Repeat
(e^x) / n(n-1) * x^n-2

So now you know that if you keep differentiating, the numerator remains unaffected. However the denominator will reduce to something like
(n)(n-1)(n-2)......1 (= say N)

Now put the limiting value of x=0
you get
1/ N


Regarding the second one, i think the left hand limit will come out to be 0. Since the domain of the function is (-inf,5] , the LHL exists as x->5-

Hope it helps !

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Isn't n(n-1)(n-2)...(3)(2)(1) = n! ?

Yeah. 1/n!
Missed that. Thanks ! ;)

Quote:

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Originally Posted by Prove It

Isn't n(n-1)(n-2)...(3)(2)(1) = n! ?

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Just btw, if we expand e^x and then see, we have
1+x+x²/2! + x³/3! + ...... xⁿ/n! + ..... to inf

so our expression becomes

(1+x+x²/2! + x³/3! + ...... xⁿ/n! + ..... to inf - (1+x) )/ xⁿ
= (x²/2! + x³/3! + ...... xⁿ/n! + ..... to inf ) / xⁿ


now for terms beyond xⁿ we need not worry as the higher powers will go to 0 as x-> 0
however if we are to find the limit by dividing the rest of the terms :-
(x²/2! + x³/3! + ...... xⁿ/n!) / xⁿ
for x-> 0
a number of undefined terms will be encountered, apart from 1/n! which seems to be coming if we can somehow deal with the lower powers !

Kindly clarify this !
thanks.
:)