The as x approaches 0 of tan^2x/x

Mathematics

1 answer:

uysha [10]3 years ago

3 0

L=Lim tan(x)^2/x x->0

Since both numerator and denominator evaluate to zero, we could apply l'Hôpital rule by taking derivatives.

d(tan^2(x))/dx=2tan(x).d(tan(x))/dx = 2tan(x)sec^2(x)

d(x)/dx = 1

L=2tan(x)sec^2(x)/1 x->0

= (2(0)/1^2)/1

=0/1

Another way using series,

We know that tan(x) = x+x^3/3+2x^5/15+.....

then tan^2(x), using binomial expansion gives

x^2+2*x^4/3+.... (we only need two terms)

and again apply l'Hôpital's rule, we have

L=d(x^2+2x^4/3+...)/d(x) = (2x+8x^3/3+...)/1

=0 as x->0

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Karo-lina-s [1.5K]

$\text{The scientific notation:}\\\\a\times10^n\\\\1\leq a$

$a.\\(5\times10^3)\times(3.5\times10^{-1})=(5\times3.5)\times(10^3\times10^{-1})\\\\=15.75\times10^{3+(-1)}=15.75\times10^2=1.575\times10\times10^2\\\\=1.575\times10^{1+2}=\boxed{1.575\times10^3}$

$b.\\(8\times10^2)\div(4\times10^3)=(8\div4)\times(10^2\div10^3)=2\times10^{2-3}\\\\=\boxed{2\times10^{-1}}$

$c.\\(6\times10^{-3})^2=6^2\times(10^{-3})^2=36\times10^{-3\cdot2}=36\times10^{-6}=3.6\times10\times10^{-6}\\\\=3.6\times10^{1+(-6)}=\boxed{3.6\times10^{-5}}$