Question

assignment 3: problem 7 (1 point) evaluate the limit $lim_{x
ightarrow0}\frac{\tan5x}{sin4x}$

Explanation:

Step1: Rewrite tangent as sine/cosine

We know that $\tan(5x)=\frac{\sin(5x)}{\cos(5x)}$, so the limit becomes $\lim_{x
ightarrow0}\frac{\sin(5x)}{\cos(5x)\sin(4x)}$.

Step2: Use the limit - formula $\lim_{u

ightarrow0}\frac{\sin u}{u} = 1$
We can rewrite the limit as $\lim_{x
ightarrow0}\frac{\sin(5x)}{5x}\times\frac{4x}{\sin(4x)}\times\frac{5x}{4x}\times\frac{1}{\cos(5x)}$.

Step3: Evaluate the limits of each part

We know that $\lim_{x
ightarrow0}\frac{\sin(5x)}{5x}=1$ and $\lim_{x
ightarrow0}\frac{\sin(4x)}{4x}=1$, and $\lim_{x
ightarrow0}\cos(5x)=\cos(0) = 1$.
The limit of the whole expression is $\lim_{x
ightarrow0}\frac{\sin(5x)}{5x}\times\frac{4x}{\sin(4x)}\times\frac{5}{4}\times\frac{1}{\cos(5x)}=1\times1\times\frac{5}{4}\times1$.

Answer:

$\frac{5}{4}$