Question

find the limit. use lhospitals rule if appropriate. if there is a more elementary method, consider using it.
lim _{x
ightarrow 0} \frac{e^{7 x}-1-7 x}{x^{2}}
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1. -/1 points

find the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using it.
lim _{x
ightarrow 0} \frac{x 2^{x}}{2^{x}-1}

Explanation:

Step1: Check if L'Hospital's Rule is applicable

When \(x = 0\), the numerator \(e^{7x}-1 - 7x=e^{0}-1-0 = 0\) and the denominator \(x^{2}=0\). So, \(\lim_{x
ightarrow0}\frac{e^{7x}-1 - 7x}{x^{2}}\) is in the \(\frac{0}{0}\) form.

Step2: Apply L'Hospital's Rule

Differentiate the numerator and the denominator.
The derivative of \(y = e^{7x}-1 - 7x\) is \(y^\prime=7e^{7x}-7\) (using the chain - rule \((e^{u})^\prime=e^{u}\cdot u^\prime\) where \(u = 7x\) and \((7x)^\prime=7\), \((-1)^\prime = 0\)).
The derivative of \(y=x^{2}\) is \(y^\prime = 2x\).
So, \(\lim_{x
ightarrow0}\frac{e^{7x}-1 - 7x}{x^{2}}=\lim_{x
ightarrow0}\frac{7e^{7x}-7}{2x}\).
When \(x = 0\), the numerator \(7e^{0}-7=0\) and the denominator \(2x = 0\). So, it is still in the \(\frac{0}{0}\) form.

Step3: Apply L'Hospital's Rule again

Differentiate the numerator and the denominator of \(\frac{7e^{7x}-7}{2x}\).
The derivative of \(y = 7e^{7x}-7\) is \(y^\prime=49e^{7x}\) (using the chain - rule \((e^{u})^\prime=e^{u}\cdot u^\prime\) where \(u = 7x\) and \((7x)^\prime=7\), \((-7)^\prime = 0\)).
The derivative of \(y = 2x\) is \(y^\prime=2\).
So, \(\lim_{x
ightarrow0}\frac{7e^{7x}-7}{2x}=\lim_{x
ightarrow0}\frac{49e^{7x}}{2}\).

Step4: Evaluate the limit

Substitute \(x = 0\) into \(\frac{49e^{7x}}{2}\).
Since \(e^{0}=1\), we have \(\frac{49e^{0}}{2}=\frac{49\times1}{2}\).

Answer:

\(\frac{49}{2}\)