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How do you find #lim (1-cosx)/x# as #x->0# using l'Hospital's Rule?

Answer 1

Elijah Abram

#0#

Note that we're in indeterminate form (#0/0#), so we can use l'Hospital's Rule.

L'hospital's Rule states that #lim_(x->c) (g(x))/(h(x)) = lim_(x->c) (g'(x))/(h'(x))#.

Therefore:

#lim_(x->0) (1- cosx)/x#

#=lim_(x->0) ((1 - cosx)')/(x')#

#=lim_(x->0) (sinx)/1#

#=sin(0)/1#

#= 0#

Hopefully this helps!

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Answer 2

Evelyn Agard

To find lim (1-cosx)/x as x approaches 0 using L'Hospital's Rule, you first need to check if the limit is in an indeterminate form, which it is (0/0). Then, you can differentiate the numerator and the denominator separately with respect to x. After differentiation, you can evaluate the limit again. Here's the process:

Given lim (1-cosx)/x as x approaches 0:

Differentiate the numerator: d(1-cosx)/dx = sin(x)
Differentiate the denominator: dx/dx = 1
Rewrite the limit with the derivatives: lim sin(x)/1 as x approaches 0
Evaluate the limit: sin(0)/1 = 0

Therefore, lim (1-cosx)/x as x approaches 0 is equal to 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find #lim (1-cosx)/x# as #x-&gt;0# using l'Hospital's Rule?

How do you find #lim (1-cosx)/x# as #x->0# using l'Hospital's Rule?