Key to "Compute the following limits using LHospitals Rule if..."

smileycellist2

Answered question

2021-10-15

Compute the following limits using LHospitals Rule if appropriate.
The limit of

lim_{x \to 1} \frac{7^{x} - 7}{x^{2} - 1}

and

lim_{x \to \infty} \frac{\tan^{- 1} x}{\frac{1}{x} - 7}

nitruraviX

Skilled2021-10-16Added 101 answers

The limit expressions are

lim_{x \to 1} \frac{7^{x} - 7}{x^{2} - 1}

and

lim_{x \to \infty} \frac{\tan^{- 1} x}{\frac{1}{x} - 7}

Complete the limits as follows.
The limit of

\frac{7^{x} - 7}{x^{2} - 1}

as x approaches 1

lim_{x \to 1} \frac{7^{x} - 7}{x^{2} - 1} = \frac{0}{0}

at x=1

lim_{x \to 1} \frac{7^{x} - 7}{x^{2} - 1} = lim_{x \to 1} \frac{7^{x} \log 7}{2 x}

lim_{x \to 1} \frac{7^{x} - 7}{x^{2} - 1} = \frac{7 \log 7}{2}

The limit

\frac{\tan^{- 1} x}{\frac{1}{x} - 7}

as x approaches

\infty

lim_{x \to \infty} \frac{\tan^{- 1} x}{\frac{1}{x} - 7} = \frac{\tan^{- 1} (\infty)}{\frac{1}{\infty} - 7}

= \frac{\frac{π}{2}}{- 7}

= - \frac{π}{14}

Jeffrey Jordon

Expert2022-07-05Added 2605 answers

Answer is given below (on video)

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