How to find the limit 5(1-cos2x)/sin2x as x approach 0 from the left ?

Question

Christopher Allers · Accepted Answer

#lim_(x->0^-) (5(1-cos2x))/(sin2x) = 0#

To assess:

#lim_(x->0^-) (5(1-cos2x))/(sin2x)#

we can divide numerator and denominator by #2x# and then substitute #t= 2x#:

#lim_(x->0^-) (5(1-cos2x))/(sin2x) = 5 lim_(x->0^-) ((1-cos2x)/(2x))/((sin2x)/(2x)) = 5 lim_(t->0^-) ((1-cost)/t)/((sint)/t)#

Now think about the well-known boundaries:

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

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

and we possess:

#lim_(x->0^-) (5(1-cos2x))/(sin2x) = (5*0)/1 = 0#

graph{sin(2x) [-10, 10, -5, 5]}/(5(1-cos(2x))

Luke Alvarez · Answer

#0.#

We have, #1-cos2x=2sin^2x, and, sin2x=2sinxcosx.#

#:." The Reqd. Limit="lim_(x to 0){5(2sin^2x)}/{2sinxcosx}#

#=5{lim_(x to 0) tanx}#

#=5(tan0)#

#=0.#

Because the Limit is real, we have

#lim_(xto0-){5(1-cos2x)}/(sin2x)=0=lim_(xto0+){5(1-cos2x)}/(sin2x)," too."#

Have fun with math!

Cameron Alexander · Answer

undefined To find the limit of $ \frac{5(1 - \cos(2x))}{\sin(2x)} $ as $ x $ approaches $ 0 $ from the left:

1. Replace $ x $ with $ 0 $ in the expression.
2. Simplify the expression.
3. Evaluate the limit.

Step 1: Substitute $ x = 0 $ into the expression:
$$ \frac{5(1 - \cos(2 \cdot 0))}{\sin(2 \cdot 0)} $$

Step 2: Simplify the expression:
$$ \frac{5(1 - \cos(0))}{\sin(0)} $$
$$ = \frac{5(1 - 1)}{0} $$
$$ = \frac{5 \cdot 0}{0} $$

Step 3: Evaluate the limit:
$$ \lim_{x \to 0^-} \frac{5(1 - \cos(2x))}{\sin(2x)} = \lim_{x \to 0^-} \frac{0}{0} $$

When you have $ \frac{0}{0} $, it indicates that the expression is indeterminate. To evaluate this type of limit, you can use L'Hôpital's Rule, which states that if the limit of the quotient of two functions approaches $ \frac{0}{0} $ or $ \frac{\infty}{\infty} $, then you can differentiate the numerator and the denominator separately and take the limit again.

$$ \lim_{x \to 0^-} \frac{5(1 - \cos(2x))}{\sin(2x)} = \lim_{x \to 0^-} \frac{5(2\sin(2x))}{2\cos(2x)} $$

$$ = \lim_{x \to 0^-} \frac{5\sin(2x)}{\cos(2x)} $$

Now, substitute $ x = 0 $ into this expression:

$$ = \frac{5\sin(0)}{\cos(0)} $$
$$ = \frac{5 \cdot 0}{1} $$
$$ = 0 $$

So, the limit of $ \frac{5(1 - \cos(2x))}{\sin(2x)} $ as $ x $ approaches $ 0 $ from the left is $ 0 $.