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How do you evaluate the limit #sinx/tanx# as x approaches #0#?

Answer 1

Aurora Alvarado

#Lt_(x->0)sinx/tanx=1#

#Lt_(x->0)sinx/tanx#

= #Lt_(x->0)sinx/(sinx/cosx)#

= #Lt_(x->0)sinx xxcosx/sinx#

= #Lt_(x->0)cosx#

= #cos0#

= #1#

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Answer 2

Isabella Ackerman

# lim_(xrarr0) sinx/tanx=1#.

We use the Standard Form of Limit :

#lim_(thetararr0)sintheta/theta=1, theta in RR#.

From the above, we have,

#lim_(thetararr0) tantheta/theta#

#=lim_(thetararr0) {sintheta/theta*1/costheta}#

#={lim_(thetararr0) sintheta/theta}*{lim_(thetararr0)1/costheta}#

#=1*(1/cos0)#

# :. lim_(thetararr0) tantheta/theta=1.#

Now, # lim_(xrarr0) sinx/tanx#

#=lim_(xrarr0) (sinx/x)(x/tanx)#

#={lim_(xrarr0) sinx/x}{lim_(xrarr0) x/tanx}#

#=1*1#

# :. lim_(xrarr0) sinx/tanx=1#.

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Answer 3

Paisley Johnson

To evaluate the limit of sinx/tanx as x approaches 0, we can simplify the expression using trigonometric identities. The identity tanx = sinx/cosx can be used to rewrite the expression as sinx/(sinx/cosx). Simplifying further, we get cosx. Therefore, the limit of sinx/tanx as x approaches 0 is equal to 1.

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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.