$$\\lim_{x \\to a} f(x)=0$$ $$\\lim_{x \\to a} g(x)=0$$ $$\\lim_{x \\to a} h(x)=1$$ $$\\lim_{x \\to a}…

$$\\lim_{x \\to a} f(x)=0$$ $$\\lim_{x \\to a} g(x)=0$$ $$\\lim_{x \\to a} h(x)=1$$ $$\\lim_{x \\to a} p(x)=\\infty$$ $$\\lim_{x \\to a} q(x)=\\infty,$$ evaluate if the following limits not are indeterminate forms. (if a limit is indeterminate, enter indeterminate.)\n(a) $$\\lim_{x \\to a}f(x)^{g(x)}$$\n(b) $$\\lim_{x \\to a}f(x)^{p(x)}$$\n(c) $$\\lim_{x \\to a}h(x)^{p(x)}$$\n(d) $$\\lim_{x \\to a}p(x)^{f(x)}$$\n(e) $$\\lim_{x \\to a}p(x)^{q(x)}$$\n(f) $$\\lim_{x \\to a} \\sqrtq(x){p(x)}$$
Answer
Explanation:
Step1: Recall indeterminate forms
Indeterminate forms include (0^{0}), (\infty^{0}), (1^{\infty}), (0\times\infty), (\frac{0}{0}), (\frac{\infty}{\infty}), (\infty-\infty)
Step2: Evaluate part (a)
We have (\lim_{x\rightarrow a}f(x) = 0) and (\lim_{x\rightarrow a}g(x)=0). The form (\lim_{x\rightarrow a}[f(x)]^{g(x)}) is of the (0^{0}) indeterminate form.
Step3: Evaluate part (b)
Since (\lim_{x\rightarrow a}f(x) = 0) and (\lim_{x\rightarrow a}p(x)=\infty), the form (\lim_{x\rightarrow a}[f(x)]^{p(x)}) is of the (0^{\infty}) form. (0^{\infty}=0) (not indeterminate)
Step4: Evaluate part (c)
Given (\lim_{x\rightarrow a}h(x) = 1) and (\lim_{x\rightarrow a}p(x)=\infty), the form (\lim_{x\rightarrow a}[h(x)]^{p(x)}) is of the (1^{\infty}) indeterminate form.
Step5: Evaluate part (d)
Since (\lim_{x\rightarrow a}p(x)=\infty) and (\lim_{x\rightarrow a}f(x) = 0), the form (\lim_{x\rightarrow a}[p(x)]^{f(x)}) is of the (\infty^{0}) indeterminate form.
Step6: Evaluate part (e)
Given (\lim_{x\rightarrow a}p(x)=\infty) and (\lim_{x\rightarrow a}q(x)=\infty), the form (\lim_{x\rightarrow a}[p(x)]^{q(x)}) is of the (\infty^{\infty}) form. (\infty^{\infty}=\infty) (not indeterminate)
Step7: Evaluate part (f)
We can rewrite (\lim_{x\rightarrow a}\sqrt[q(x)]{p(x)}=\lim_{x\rightarrow a}[p(x)]^{\frac{1}{q(x)}}). Since (\lim_{x\rightarrow a}p(x)=\infty) and (\lim_{x\rightarrow a}q(x)=\infty), then (\lim_{x\rightarrow a}\frac{1}{q(x)} = 0). The form (\lim_{x\rightarrow a}[p(x)]^{\frac{1}{q(x)}}) is of the (\infty^{0}) indeterminate form.
Answer:
(a) INDETERMINATE (b) (0) (c) INDETERMINATE (d) INDETERMINATE (e) (\infty) (f) INDETERMINATE