## On what interval is the function $$f(x)=-x^3+2x^2+5$$ increasing and concave downward?

### Solution.

• We need to find out for which $$x$$ is $$f'(x)>0$$ and $$f''(x)<0$$.
• We have $$f'(x)=-3x^2+4x$$. This is a quadratic function with negative leading coefficient and roots $$x=0,\frac43$$. Therefore, it is positive when $$0<x<\frac43$$.
• We have $$f''(x)=-6x+4$$. This is a linear function with negative leading coefficient has root $$x=\frac23$$. Therefore, it is negative when $$x>\frac23$$.
• Therefore, $$f(x)$$ is increasing and concave downward on $$(\frac23,\frac43$$.)

## On what interval is the function $$f(x)=-x^3+4x^2-5x$$ decreasing and concave downward?

### Solution.

• We need to find out for which $$x$$ is $$f'(x)<0$$ and $$f''(x)<0$$.
• We have $$f'(x)=-3x^2+8x-5$$. This is a quadratic function with negative leading coefficient and roots $$x=1,\frac53$$. Therefore, it is negative when $$x<1$$ or $$x>\frac53$$.
• We have $$f''(x)=-6x+8$$. This is a linear function with negative leading coefficient and root $$x=\frac43$$. Therefore, it is negative when $$x>\frac43$$.
• Therefore, $$f(x)$$ is decreasing and concave downward when $$x>\frac53$$.

## Find the limit $$\lim_{x\to0^+}(\sin x)^{\sin x}$$.

### Solution.

• This is an indeterminate power of type $$0^0$$. Therefore, we need to find $$\lim_{x\to0^+}(\sin x)\ln(\sin x)$$.
• That in turn is an indeterminate product of type $$0\cdot\infty$$, therefore we can apply the L'Hospital rule: $\lim_{x\to0^+}(\sin x)\ln(\sin x)=\lim_{x\to0^+}\frac{\ln(\sin x)}{(\sin x)^{-1}}=\lim_{x\to0^+}\frac{\frac{\cos x}{\sin x}}{\frac{-\cos x}{(\sin x)^2}}=\lim_{x\to0^+}(-\sin x)=0.$
• This shows that $$\lim_{x\to0^+}(\sin x)^{\sin x}=1$$.

## Find the limit $$\lim_{x\to0^+}(\sin x)^{\tan x}$$.

### Solution.

• This is an indeterminate power of type $$0^0$$. Therefore, we need to find $$\lim_{x\to0^+}(\tan x)\ln(\sin x)$$.
• That in turn is an indeterminate product of type $$0\cdot\infty$$, therefore we can apply the L'Hospital rule: $\lim_{x\to0^+}(\tan x)\ln(\sin x)=\lim_{x\to0^+}\frac{\ln(\sin x)}{(\tan x)^{-1}}=\lim_{x\to0^+}\frac{\frac{\cos x}{\sin x}}{\frac{-1}{(\sin x)^2}}=\lim_{x\to0^+}(-(\sin x)(\cos x))=0.$
• This shows that $$\lim_{x\to0^+}(\sin x)^{\tan x}=1$$.

## Find the limit $$\lim_{x\to0^+}(\tan x)^{\sin x}$$.

### Solution.

• This is an indeterminate power of type $$0^0$$. Therefore, we need to find $$\lim_{x\to0^+}(\sin x)\ln(\tan x)$$.
• That in turn is an indeterminate product of type $$0\cdot\infty$$, therefore we can apply the L'Hospital rule: $\lim_{x\to0^+}(\sin x)\ln(\tan x)=\lim_{x\to0^+}\frac{\ln(\tan x)}{(\sin x)^{-1}}=\lim_{x\to0^+}\frac{\frac{(\sec x)^2}{\tan x}}{\frac{-\cos x}{(\sin x)^2}} =\lim_{x\to0^+}\frac{\frac{\frac{1}{(\cos x)^2}}{\frac{\sin x}{\cos x}}}{\frac{-\cos x}{(\sin x)^2}}=\lim_{x\to0^+}-\frac{\sin x}{(\cos x)^2}=0.$
• This shows that $$\lim_{x\to0^+}(\tan x)^{\sin x}=1$$.

## Find the limit $$\lim_{x\to0^+}(\tan x)^{\tan x}$$.

### Solution.

• This is an indeterminate power of type $$0^0$$. Therefore, we need to find $$\lim_{x\to0^+}(\tan x)\ln(\tan x)$$.
• That in turn is an indeterminate product of type $$0\cdot\infty$$, therefore we can apply the L'Hospital rule: $\lim_{x\to0^+}(\tan x)\ln(\tan x)=\lim_{x\to0^+}\frac{\ln(\tan x)}{(\tan x)^{-1}}=\lim_{x\to0^+}\frac{\frac{(\sec x)^2}{\tan x}}{\frac{-1}{(\sin x)^2}} =\lim_{x\to0^+}\frac{\frac{\frac{1}{(\cos x)^2}}{\frac{\sin x}{\cos x}}}{\frac{-1}{(\sin x)^2}}=\lim_{x\to0^+}-\frac{\sin x}{\cos x}=0.$
• This shows that $$\lim_{x\to0^+}(\tan x)^{\tan x}=1$$.