Question

find the limit.
lim(x→0) sin(x^6)/x
0 exactly!

1. -/1 points

find the limit.
lim(x→π/4) (5 - 5 tan(x))/(sin(x) - cos(x))

Explanation:

Step1: Rewrite the expression

We know that $\tan(x)=\frac{\sin(x)}{\cos(x)}$, so the given limit $\lim_{x
ightarrow\frac{\pi}{4}}\frac{5 - 5\tan(x)}{\sin(x)-\cos(x)}=\lim_{x
ightarrow\frac{\pi}{4}}\frac{5(1 - \frac{\sin(x)}{\cos(x)})}{\sin(x)-\cos(x)}=\lim_{x
ightarrow\frac{\pi}{4}}\frac{5(\cos(x)-\sin(x))}{\cos(x)(\sin(x)-\cos(x))}$.

Step2: Simplify the expression

$\lim_{x
ightarrow\frac{\pi}{4}}\frac{5(\cos(x)-\sin(x))}{\cos(x)(\sin(x)-\cos(x))}=\lim_{x
ightarrow\frac{\pi}{4}}\frac{- 5(\sin(x)-\cos(x))}{\cos(x)(\sin(x)-\cos(x))}$. Cancel out the non - zero factor $\sin(x)-\cos(x)$ (since $x
ightarrow\frac{\pi}{4}$ but $x
eq\frac{\pi}{4}$), we get $\lim_{x
ightarrow\frac{\pi}{4}}\frac{-5}{\cos(x)}$.

Step3: Evaluate the limit

Substitute $x = \frac{\pi}{4}$ into $\frac{-5}{\cos(x)}$. Since $\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$, then $\lim_{x
ightarrow\frac{\pi}{4}}\frac{-5}{\cos(x)}=\frac{-5}{\frac{\sqrt{2}}{2}}=-5\sqrt{2}$.

Answer:

$-5\sqrt{2}$