1. (a) use the squeeze theorem to calculate the following limits. i) $lim_{x\rightarrow - 2^{-}}sqrt{x +…

1. (a) use the squeeze theorem to calculate the following limits. i) $lim_{x\rightarrow - 2^{-}}sqrt{x + 2}sinleft(\frac{1}{x + 2}\right)$ ii) $lim_{x\rightarrowinfty}\frac{1+sin(x)}{x^{2}}$ (b) suppose that $f(x)$ is a function that satisfies $-3x + 1leq f(x)leq x^{2}-3$ for all $x>1$. calculate $lim_{x\rightarrow1^{+}}f(x)$. justify your answer using all relevant theorems.

1. (a) use the squeeze theorem to calculate the following limits. i) $lim_{x\rightarrow - 2^{-}}sqrt{x + 2}sinleft(\frac{1}{x + 2}\right)$ ii) $lim_{x\rightarrowinfty}\frac{1+sin(x)}{x^{2}}$ (b) suppose that $f(x)$ is a function that satisfies $-3x + 1leq f(x)leq x^{2}-3$ for all $x>1$. calculate $lim_{x\rightarrow1^{+}}f(x)$. justify your answer using all relevant theorems.

Answer

Explanation:

Step1: Recall the Squeeze Theorem

The Squeeze Theorem states that if (g(x)\leq h(x)\leq k(x)) for all (x) in some open - interval containing (a) (except possibly at (x = a)) and (\lim_{x\rightarrow a}g(x)=\lim_{x\rightarrow a}k(x)=L), then (\lim_{x\rightarrow a}h(x)=L).

Step2: Solve (a)(i)

We know that (- 1\leq\sin\left(\frac{1}{x + 2}\right)\leq1) for (x\neq - 2). Then (-\sqrt{x + 2}\leq\sqrt{x + 2}\sin\left(\frac{1}{x + 2}\right)\leq\sqrt{x + 2}). We calculate (\lim_{x\rightarrow - 2^{-}}(-\sqrt{x + 2}) = 0) and (\lim_{x\rightarrow - 2^{-}}\sqrt{x + 2}=0). By the Squeeze Theorem, (\lim_{x\rightarrow - 2^{-}}\sqrt{x + 2}\sin\left(\frac{1}{x + 2}\right)=0).

Step3: Solve (a)(ii)

We know that (-1\leq\sin(x)\leq1). Then (0\leq1+\sin(x)\leq2). So (\frac{1}{x^{2}}\cdot0\leq\frac{1+\sin(x)}{x^{2}}\leq\frac{1}{x^{2}}\cdot2). We calculate (\lim_{x\rightarrow\infty}0 = 0) and (\lim_{x\rightarrow\infty}\frac{2}{x^{2}}=0). By the Squeeze Theorem, (\lim_{x\rightarrow\infty}\frac{1+\sin(x)}{x^{2}}=0).

Step4: Solve (b)

We are given (-3x + 1\leq f(x)\leq x^{2}-3) for all (x>1). We calculate (\lim_{x\rightarrow1^{+}}(-3x + 1)=-3\times1 + 1=-2) and (\lim_{x\rightarrow1^{+}}(x^{2}-3)=1^{2}-3=-2). By the Squeeze Theorem, (\lim_{x\rightarrow1^{+}}f(x)=-2).

Answer:

(a)(i) (0) (a)(ii) (0) (b) (-2)