Question

$\lim_{x\to3}\frac{e^{x-3}-x+1}{x^{2}-\log(x-2)}$ is equal to

• A. $\frac{-1}{3}$
• B. $\frac{-2}{9}$
• C. $\frac{-1}{2}$
• D. $\frac{-1}{4}$
• E. $\frac{-1}{9}$

Hint:

Calculus Tip: Do not blindly jump into L'Hôpital's Rule just because you see an $e^x$ or a logarithm in a limit! Always test direct substitution first; it takes 5 seconds and often solves the problem instantly.

Answer 1

Answer: (E)

Concept:
When evaluating a limit, always attempt direct substitution first. If substituting the limit value directly into the function yields a defined real number (not an indeterminate form like $0/0$ or $\infty/\infty$), then that number is the limit.

Step 1: State the limit expression.
We need to evaluate: $$L = \lim_{x\rightarrow3}\frac{e^{x-3}-x+1}{x^{2}-\log(x-2)}$$

Step 2: Substitute $x = 3$ into the numerator.
Evaluate the top half of the fraction: $$\text{Numerator} = e^{3-3} - 3 + 1$$ $$\text{Numerator} = e^0 - 2$$ Since $e^0 = 1$: $$\text{Numerator} = 1 - 2 = -1$$

Step 3: Substitute $x = 3$ into the denominator.
Evaluate the bottom half of the fraction: $$\text{Denominator} = (3)^2 - \log(3 - 2)$$ $$\text{Denominator} = 9 - \log(1)$$ Since the logarithm of 1 (in any base) is exactly 0: $$\text{Denominator} = 9 - 0 = 9$$

Step 4: Check for indeterminate forms.
The result of our direct substitution is $\frac{-1}{9}$. Because the denominator is non-zero and the numerator is defined, this is not an indeterminate form. No advanced limit techniques (like L'Hôpital's rule) are required.

Step 5: State the final limit.
Combine the evaluated numerator and denominator: $$L = \frac{-1}{9}$$ Hence the correct answer is (E) $\frac{-1{9}$}.