given that\nlim_{x\\to1} f(x)=1 lim_{x\\to1} g(x)= - 5 lim_{x\\to1} h(x)=0,\nfind the limits, if they exist…

Explanation:

Step1: Use limit - sum rule

The limit of a sum $\lim_{x\rightarrow a}[f(x)+g(x)]=\lim_{x\rightarrow a}f(x)+\lim_{x\rightarrow a}g(x)$ and the constant - multiple rule $\lim_{x\rightarrow a}[cf(x)] = c\lim_{x\rightarrow a}f(x)$. For $\lim_{x\rightarrow 1}[f(x)+4g(x)]$, we have $\lim_{x\rightarrow 1}[f(x)+4g(x)]=\lim_{x\rightarrow 1}f(x)+4\lim_{x\rightarrow 1}g(x)$. Substitute $\lim_{x\rightarrow 1}f(x) = 1$ and $\lim_{x\rightarrow 1}g(x)=-5$: $1 + 4\times(-5)=1-20=-19$.

Step2: Use power - rule for limits

The power - rule states that $\lim_{x\rightarrow a}[f(x)]^n=[\lim_{x\rightarrow a}f(x)]^n$. For $\lim_{x\rightarrow 1}[g(x)]^3$, substitute $\lim_{x\rightarrow 1}g(x)=-5$: $(-5)^3=-125$.

Step3: Use root - rule for limits

The root - rule for limits is $\lim_{x\rightarrow a}\sqrt{f(x)}=\sqrt{\lim_{x\rightarrow a}f(x)}$ (when $\lim_{x\rightarrow a}f(x)\geq0$ for even roots). For $\lim_{x\rightarrow 1}\sqrt{f(x)}$, substitute $\lim_{x\rightarrow 1}f(x) = 1$: $\sqrt{1}=1$.

Step4: Use quotient - rule for limits

The quotient - rule for limits is $\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x\rightarrow a}f(x)}{\lim_{x\rightarrow a}g(x)}$ (when $\lim_{x\rightarrow a}g(x)\neq0$). For $\lim_{x\rightarrow 1}\frac{5f(x)}{g(x)}$, we have $\lim_{x\rightarrow 1}\frac{5f(x)}{g(x)}=\frac{5\lim_{x\rightarrow 1}f(x)}{\lim_{x\rightarrow 1}g(x)}$. Substitute $\lim_{x\rightarrow 1}f(x) = 1$ and $\lim_{x\rightarrow 1}g(x)=-5$: $\frac{5\times1}{-5}=-1$.

Step5: Analyze the quotient

For $\lim_{x\rightarrow 1}\frac{g(x)}{h(x)}$, since $\lim_{x\rightarrow 1}g(x)=-5$ and $\lim_{x\rightarrow 1}h(x)=0$, the limit is of the form $\frac{-5}{0}$, so $\lim_{x\rightarrow 1}\frac{g(x)}{h(x)}=\text{DNE}$.

Step6: Use product - rule for limits

The product - rule for limits is $\lim_{x\rightarrow a}[f(x)g(x)]=\lim_{x\rightarrow a}f(x)\cdot\lim_{x\rightarrow a}g(x)$. For $\lim_{x\rightarrow 1}\frac{g(x)h(x)}{f(x)}$, we have $\lim_{x\rightarrow 1}\frac{g(x)h(x)}{f(x)}=\frac{\lim_{x\rightarrow 1}g(x)\cdot\lim_{x\rightarrow 1}h(x)}{\lim_{x\rightarrow 1}f(x)}$. Substitute $\lim_{x\rightarrow 1}f(x) = 1$, $\lim_{x\rightarrow 1}g(x)=-5$ and $\lim_{x\rightarrow 1}h(x)=0$: $\frac{-5\times0}{1}=0$.

Answer:

(a) - 19 (b) - 125 (c) 1 (d) - 1 (e) DNE (f) 0