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2 Answers
Jan 4, 2018

lim_(x->0) (sin 4x)/(tan 5x) = 4/5

Explanation:

Quick method

Note that:

lim_(t->0) sin t / t = 1

lim_(t->0) tan t / t = 1

That is, both sin t and tan t behave like t for small values of t.

So:

lim_(x->0) (sin 4x) / (tan 5x) = lim_(x->0) (4x)/(5x) = 4/5

L'Hôpital's rule method

Note that:

lim_(x->0) sin 4x = 0

lim_(x->0) tan 5x = 0

So the requested limit is an indeterminate form 0/0 and hence L'Hôpital's rule applies.

lim_(x->0) (sin 4x)/(tan 5x) = lim_(x->0) (d/(dx) sin 4x)/(d/(dx) tan 5x)

color(white)(lim_(x->0) (sin 4x)/(tan 5x)) = lim_(x->0) ((d/(dx) 4x)(cos 4x))/((d/(dx) 5x)(sec^2 5x))

color(white)(lim_(x->0) (sin 4x)/(tan 5x)) = lim_(x->0) (4 cos 4x)/(5 sec^2 5x)

color(white)(lim_(x->0) (sin 4x)/(tan 5x)) = (4 * 1)/(5 * 1)

color(white)(lim_(x->0) (sin 4x)/(tan 5x)) = 4/5

Jan 4, 2018

lim_(x->0)sin(4x)/tan(5x)=4/5

Explanation:

lim_(x->0)sin(4x)/tan(5x)

Since tan(theta)=sin(theta)/cos(theta), we have
=lim_(x->0)(sin(4x)cos(5x))/sin(5x)
=lim_(x->0)sin(4x)/sin(5x)

Now, arrange to
=lim_(x->0)4/5*sin(4x)/(4x)*(5x)/sin(5x)

If the factors are canceled out, this will be equal to the previous expression. The reason for this arrangement is to exploit the identity lim_(x->0)sin(x)/x=1.

Then, we have
=lim_(x->0)4/5*lim_(x->0)sin(4x)/(4x)*lim_(x->0)(5x)/sin(5x)

As x->0, 4x->0 and 5x->0. Therefore, the above expression becomes
=lim_(x->0)4/5*lim_(4x->0)sin(4x)/(4x)*lim_(5x->0)(5x)/sin(5x)

Now we can use lim_(x->0)sin(x)/x=lim_(x->0)x/sin(x)=1 to evaluate the individual limits:
=4/5*1*1
=4/5

We can verify this limit with a graph:
graph{sin(4x)/tan(5x) [-1,1,-1,1]}

As seen, as x->0, sin(4x)/tan(5x)->4/5.