Step 1

\(\lim\arctan(e^{x})=\arctan(e^{\infty})\)

\(=\arctan(\infty)=\frac{\pi}{2}\)

\(\lim\arctan(e^{x})=\arctan(e^{\infty})\)

\(=\arctan(\infty)=\frac{\pi}{2}\)

asked 2021-10-20

Find the limit, if it exists, or show that the limits does not exist.

\(\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({3}\pi,{2}\pi\right)}}}{x}{y}{\cos{{\left({x}-{2}{y}\right)}}}\).

\(\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({3}\pi,{2}\pi\right)}}}{x}{y}{\cos{{\left({x}-{2}{y}\right)}}}\).

asked 2021-11-11

Find the limit or show that it does not exist.

\(\displaystyle\lim_{{{x}\to\infty}}={\frac{{{1}-{x}^{{2}}}}{{{x}^{{3}}-{x}+{1}}}}\)

\(\displaystyle\lim_{{{x}\to\infty}}={\frac{{{1}-{x}^{{2}}}}{{{x}^{{3}}-{x}+{1}}}}\)

asked 2021-11-06

Find the limits. if the limit does not exist, state it. In each part show your calculations and circle the answer.

\(\displaystyle{\frac{{{x}^{{2}}-{4}{x}+{4}}}{{{x}^{{2}}+{x}-{6}}}}\)

\(\displaystyle{\frac{{{x}^{{2}}-{4}{x}+{4}}}{{{x}^{{2}}+{x}-{6}}}}\)

asked 2021-10-26

Find the limit.If limit does not exist state it. In each part show your calculations and circle the answer

\(\displaystyle\lim_{{{x}\to\infty}}{\frac{{{5}{x}^{{2}}+{7}}}{{{3}{x}^{{2}}-{x}}}}\)

\(\displaystyle\lim_{{{x}\to\infty}}{\frac{{{5}{x}^{{2}}+{7}}}{{{3}{x}^{{2}}-{x}}}}\)

asked 2021-08-06

To find:

The value of \(\displaystyle\lim_{{{x}\rightarrow{0}}}{\left({\frac{{{4}^{{{3}{x}}}-{4}^{{x}}}}{{{4}^{{x}}-{1}}}}\right)}\) if it exists. If limit does not exist , then find out if one-sided limit exists.

The value of \(\displaystyle\lim_{{{x}\rightarrow{0}}}{\left({\frac{{{4}^{{{3}{x}}}-{4}^{{x}}}}{{{4}^{{x}}-{1}}}}\right)}\) if it exists. If limit does not exist , then find out if one-sided limit exists.

asked 2021-10-19

Find the limit if it exists. If it does hot cxíst explain why Ih the case of ihfinite limits, malk suve that you check both the left and right limit at the point in the question.

\(\displaystyle\lim_{{{x}\to{1}}}{\frac{{{\left({x}^{{2}}-{1}\right)}{e}^{{{\frac{{{1}}}{{{x}-{1}}}}}}}}{{{\left({x}-{1}\right)}}}}\)

\(\displaystyle\lim_{{{x}\to{1}}}{\frac{{{\left({x}^{{2}}-{1}\right)}{e}^{{{\frac{{{1}}}{{{x}-{1}}}}}}}}{{{\left({x}-{1}\right)}}}}\)

asked 2021-11-14

Find the limit (if it exists).

\(\displaystyle\lim_{{{x}\to{7}^{+}}}{\ln{{\left({x}-{7}\right)}}}\)

\(\displaystyle\lim_{{{x}\to{7}^{+}}}{\ln{{\left({x}-{7}\right)}}}\)