question 5 (1 point)\n$limlimits_{x\toinfty}7xe^{1/x}-7x=$ (enter a number for your answer.)\nquestion 6 (1…

Explanation:

Step1: Rewrite the expression

We know that (\lim_{x\rightarrow\infty}7xe^{\frac{1}{x}} - 7x=7\lim_{x\rightarrow\infty}x\left(e^{\frac{1}{x}} - 1\right)). Let (t=\frac{1}{x}), then as (x\rightarrow\infty), (t\rightarrow0). So the limit becomes (7\lim_{t\rightarrow0}\frac{e^{t}-1}{t}).

Step2: Use L - H rule

By L - H rule ((\lim_{t\rightarrow a}\frac{f(t)}{g(t)}=\lim_{t\rightarrow a}\frac{f^{\prime}(t)}{g^{\prime}(t)}) when (\lim_{t\rightarrow a}f(t)=\lim_{t\rightarrow a}g(t) = 0)), for (y = e^{t}-1) and (z=t), (y^{\prime}=e^{t}) and (z^{\prime}=1). Then (\lim_{t\rightarrow0}\frac{e^{t}-1}{t}=\lim_{t\rightarrow0}\frac{e^{t}}{1}).

Step3: Evaluate the limit

Substitute (t = 0) into (\frac{e^{t}}{1}), we get (\frac{e^{0}}{1}=1). So (7\lim_{t\rightarrow0}\frac{e^{t}-1}{t}=7\times1 = 7).

For (\lim_{x\rightarrow\infty}(1-\frac{6}{x})^{x}), we use the formula (\lim_{x\rightarrow\infty}(1+\frac{a}{x})^{x}=e^{a}). Here (a=-6), so (\lim_{x\rightarrow\infty}(1-\frac{6}{x})^{x}=e^{-6}).

Answer:

Question 5: (7) Question 6: (e^{-6})