lim(h)-->0 ((sqrt(1+h)-1) / h?

Question

I'm suposed to evaluate the limit.

lim(h)-->0 ((sqrt(1+h)-1) / h

I keep multiplying by the conjugate of the numerator, (conjugate/conjugate) ,
but then the denominatore becomes h*sqrt(1+h) + h


Like this:

h+2
--------
h*sqrt(1+h) + h

So I still have a denominator aproaching 0 and I can't cancel anything.

The original equations numerator does not factor.

Answer 1

We have to evaluate, limit {sqrt(1+h) -1} / h
h--->0

Multiplying the conjugate of the numerator in both the numerator(N) & the denominator(D), we get:-

= limit [{sqrt(1+h) -1}*{sqrt(1+h) +1}] / [h*{sqrt(1+h) +1}]
h--->0

= limit [1+h - 1] / [h*{sqrt(1+h) +1}]
h--->0

= limit h / [h*{sqrt(1+h) + 1}]
h--->0

= limit 1/{sqrt(1+h) + 1} [Cancelling the 'h' in both N & D]
h--->0

= 1/{sqrt(1+0) +1}

=1/(1 + 1)

= 1/2

This is the answer to your problem. By the way, did you notice your mistake; your mistake was that you multiplied the 'h' in the denominator with the conjugate, which is why you were not able to cancel it with the 'h' in the numerator. And, please don't use L' Hospital's Rule (as was advised to you by some other answerer) in these sort of 'simple' problems; in fact, if you use this rule in these problems, the Teacher might even cut marks.

Answer 2

a thanks to do it extremely is to graph the function y = (tan x - x)/(x - sin x) and spot how the graph behaves as x thoughts 0 from the finest. If it extremely is complete you'll locate the fee of the function thoughts 2, as a effect the Lim (x -> 0+) (tan x - x)/(x - sin x) = 2

Answer 3

You need to use L'hopital's Rule.. It goes like this..

lim(h)--> 0 f(h)/g(h) = lim(h)-->0 f'(h)/g'(h)

So you let f(h) = sqrt(1+h) - 1
g(h) = h

Differentiating, we get,
f'(h) = 1/(2*sqrt(1+h))
g'(h) = 1

So f'(h)/g'(h) = 1/(2*sqrt(1+h))

So by L'Hopital's Rule,

lim(h)-->0 ((sqrt(1+h)-1) / h = lim(h)-->0 1/(2*sqrt(1+h))

= 1/2 (Since h tends to 0)


Hope that helps. =)

Answer 4

(2*h+2h+1)-1/h so 2*h-2h-1-1/h so 2*h-2h-h=(h+2)h-/h so it is -(h+2)=h-2-