Question

limit of tan7x/sin5x as x approaches 0. please show steps?

Answer 1

\[\lim_{x\to0}\frac{\tan7x}{\sin5x}=\lim_{x\to0}\frac{\sin7x}{1}\cdot\frac{1}{\sin5x}\cdot\frac{1}{\cos7x}\]
Recall that
\[\lim_{x\to0}\frac{\sin ax}{ax}=\lim_{x\to0}\frac{ax}{\sin ax}=1~~\text{for }a>0\]
and that the limit of products is equivalent to the product of the limits:
\[\lim_{x\to c}f(x)g(x)=\left(\lim_{x\to c}f(x)\right)\left(\lim_{x\to c}g(x)\right)\]
How can you rewrite the first line so that it agrees with the known limits? Try multiplying by \(\dfrac{7x}{7x}\) and \(\dfrac{5x}{5x}\):
\[\lim_{x\to0}\frac{\sin7x}{1}\cdot\frac{1}{\sin5x}\cdot\frac{1}{\cos7x}\cdot\frac{7x}{7x}\cdot\frac{5x}{5x}\]
Rearranging a bit, you have
\[\lim_{x\to0}\frac{\sin7x}{7x}\cdot\frac{5x}{\sin5x}\cdot\frac{1}{\cos7x}\cdot\frac{7x}{5x}\]
Then, splitting up the limit gives you
\[\left(\lim_{x\to0}\frac{\sin7x}{7x}\right)\left(\lim_{x\to0}\frac{5x}{\sin5x}\right)\left(\lim_{x\to0}\frac{1}{\cos7x}\right)\left(\lim_{x\to0}\frac{7x}{5x}\right)\]

Answer 2

nifty :)

Answer 3

thank you so much!

Answer 4

you're welcome