find the limit. use lhospitals rule if appropriate. if there is a more elementary method, consider using…

find the limit. use lhospitals rule if appropriate. if there is a more elementary method, consider using it.\n lim _{x \rightarrow 0} \frac{e^{7 x}-1-7 x}{x^{2}} \nresources\n13. -/1 points\nfind the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using it.\n lim _{x \rightarrow 0} \frac{x 2^{x}}{2^{x}-1}

find the limit. use lhospitals rule if appropriate. if there is a more elementary method, consider using it.\n lim _{x \rightarrow 0} \frac{e^{7 x}-1-7 x}{x^{2}} \nresources\n13. -/1 points\nfind the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using it.\n lim _{x \rightarrow 0} \frac{x 2^{x}}{2^{x}-1}

Answer

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\rightarrow0}\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\rightarrow0}\frac{e^{7x}-1 - 7x}{x^{2}}=\lim_{x\rightarrow0}\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\rightarrow0}\frac{7e^{7x}-7}{2x}=\lim_{x\rightarrow0}\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})