Question

find the limit.
lim_{x
ightarrow0}\frac{sin(x^{6})}{x}
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1. - / 1 points

find the limit.
lim_{x
ightarrowpi/4}\frac{5 - 5\tan(x)}{sin(x)-cos(x)}
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Explanation:

Step1: Use the limit - rule $\lim_{u

ightarrow0}\frac{\sin u}{u}=1$
Let $u = x^{6}$. As $x
ightarrow0$, $u=x^{6}
ightarrow0$. So, $\lim_{x
ightarrow0}\frac{\sin(x^{6})}{x}=\lim_{x
ightarrow0}\frac{\sin(x^{6})}{x^{6}}\cdot x^{5}$. Since $\lim_{u
ightarrow0}\frac{\sin u}{u} = 1$ (with $u = x^{6}$), we have $\lim_{x
ightarrow0}\frac{\sin(x^{6})}{x^{6}}\cdot x^{5}=1\times0 = 0$.

Step2: For $\lim_{x

ightarrow\frac{\pi}{4}}\frac{5 - 5\tan(x)}{\sin(x)-\cos(x)}$
First, factor out 5 from the numerator: $\lim_{x
ightarrow\frac{\pi}{4}}\frac{5(1 - \tan(x))}{\sin(x)-\cos(x)}$. Recall that $\tan(x)=\frac{\sin(x)}{\cos(x)}$, so the expression becomes $\lim_{x
ightarrow\frac{\pi}{4}}\frac{5(1-\frac{\sin(x)}{\cos(x)})}{\sin(x)-\cos(x)}=\lim_{x
ightarrow\frac{\pi}{4}}\frac{5(\frac{\cos(x)-\sin(x)}{\cos(x)})}{\sin(x)-\cos(x)}$. Then, $\lim_{x
ightarrow\frac{\pi}{4}}\frac{5(\cos(x)-\sin(x))}{\cos(x)(\sin(x)-\cos(x))}=\lim_{x
ightarrow\frac{\pi}{4}}\frac{- 5(\sin(x)-\cos(x))}{\cos(x)(\sin(x)-\cos(x))}$. Cancel out $\sin(x)-\cos(x)$ (since $x
ightarrow\frac{\pi}{4}$ and $\sin(x)-\cos(x)
eq0$ when $x
eq\frac{\pi}{4}$), we get $\lim_{x
ightarrow\frac{\pi}{4}}\frac{-5}{\cos(x)}$. Substitute $x = \frac{\pi}{4}$, and $\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$, so $\lim_{x
ightarrow\frac{\pi}{4}}\frac{-5}{\cos(x)}=-5\sqrt{2}$.

Answer:

1. $0$
2. $- 5\sqrt{2}$