L'Hospital's Rule - Calculus 1

1. Evaluate $\lim_{x \to \pi} \frac{\sin(x)}{x}$.
   Answer: $0$ This limit is not an indeterminate form. We can evaluate it by direct substitution: $\lim_{x \to \pi} \frac{\sin(x)}{x} = \frac{\sin(\pi)}{\pi} = \frac{0}{\pi} = 0.$ L'Hospital's Rule does not apply because the form is $\frac{0}{\pi}$, not $\frac{0}{0}$ or $\frac{\infty}{\infty}$. A common mistake is to mechanically apply L'Hospital's Rule, which would lead to the incorrect result $\lim_{x \to \pi} \frac{\cos(x)}{1} = \cos(\pi) = -1$.
2. Evaluate the limit: $L = \lim_{x \to 0} \frac{e^{2x} - e^{-2x} - 4x}{x - \sin(x)}$
   Answer: $L = 16$ The limit is of the indeterminate form $\frac{0}{0}$. Applying L'Hospital's Rule: $\lim_{x \to 0} \frac{2e^{2x} + 2e^{-2x} - 4}{1 - \cos(x)}$ This is still of the form $\frac{0}{0}$. Applying L'Hospital's Rule a second time: $\lim_{x \to 0} \frac{4e^{2x} - 4e^{-2x}}{\sin(x)}$ This is still of the form $\frac{0}{0}$. Applying L'Hospital's Rule a third time: $\lim_{x \to 0} \frac{8e^{2x} + 8e^{-2x}}{\cos(x)} = \frac{8e^0 + 8e^0}{\cos(0)} = \frac{8+8}{1} = 16.$
3. Let $f(x) = \begin{cases} 2\sin(x) & \text{if } x \ge 0 \\ a(e^x - 1) & \text{if } x < 0 \end{cases}$. Let $g(x) = \int_0^x f(t) \, dt$. If $f(x)$ is differentiable at $x=0$, what is the value of $\lim_{x \to 0} \frac{g(x)}{x^2}$?
   Answer: $1$ For $f(x)$ to be differentiable at $x=0$, it must first be continuous. Continuity requires $\lim_{x\to 0^+} f(x) = \lim_{x\to 0^-} f(x)$, which means $2\sin(0) = a(e^0-1)$, or $0=0$. This is always true. For differentiability, the left and right derivatives must be equal at $x=0$. $f'(x) = \begin{cases} 2\cos(x) & \text{if } x > 0 \\ ae^x & \text{if } x < 0 \end{cases}$. At $x=0$, we need $2\cos(0) = ae^0$, which implies $a=2$. So for $f(x)$ to be differentiable, $a$ must be 2. Now we evaluate the limit $\lim_{x \to 0} \frac{g(x)}{x^2}$. This is of the form $\frac{0}{0}$. Using L'Hospital's Rule and the FTC: $\lim_{x \to 0} \frac{g'(x)}{2x} = \lim_{x \to 0} \frac{f(x)}{2x}$ To evaluate this limit, we check from the left and right. Right-hand limit: $\lim_{x \to 0^+} \frac{2\sin(x)}{2x} = \lim_{x \to 0^+} \frac{\sin(x)}{x} = 1$. Left-hand limit (with $a=2$): $\lim_{x \to 0^-} \frac{2(e^x-1)}{2x} = \lim_{x \to 0^-} \frac{e^x-1}{x}$. This is of the form $\frac{0}{0}$, so we use L'Hospital's Rule again: $\lim_{x \to 0^-} \frac{e^x}{1} = 1$. Since the left and right limits are both 1, the overall limit is 1.
4. Let $f$ be a twice-differentiable function. If $\lim_{x \to 0} \frac{f(x) - 5x}{x^2} = 3$, and $f(0)=0$, what is the value of $f''(0)$?
   Answer: $6$ The expression $\frac{f(x) - 5x}{x^2}$ has the indeterminate form $\frac{0}{0}$ as $x \to 0$ since $f(0)=0$. We can apply L'Hospital's Rule: $\lim_{x \to 0} \frac{f'(x) - 5}{2x} = 3$ Since the denominator approaches 0, for the limit to be a finite number (3), the numerator must also approach 0. Therefore, $\lim_{x \to 0} (f'(x) - 5) = 0$, which implies $f'(0) - 5 = 0$, so $f'(0) = 5$. Now the limit expression $\lim_{x \to 0} \frac{f'(x) - 5}{2x}$ is again of the indeterminate form $\frac{0}{0}$. We can apply L'Hospital's Rule a second time: $\lim_{x \to 0} \frac{f''(x)}{2} = 3$ Since $f$ is twice-differentiable, $f''$ is continuous at $x=0$. Thus, we can substitute $x=0$: $\frac{f''(0)}{2} = 3 \implies f''(0) = 6.$
5. Evaluate $\lim_{x \to 1} \left( \frac{1}{\ln x} - \frac{1}{x-1} \right)$.
   Answer: $\frac{1}{2}$ The limit is of the indeterminate form $\infty - \infty$. First, combine the fractions: $\lim_{x \to 1} \frac{x-1 - \ln x}{(\ln x)(x-1)}$ As $x \to 1$, this becomes the indeterminate form $\frac{0}{0}$. Applying L'Hospital's Rule: $\lim_{x \to 1} \frac{\frac{d}{dx}(x-1 - \ln x)}{\frac{d}{dx}((\ln x)(x-1))} = \lim_{x \to 1} \frac{1 - \frac{1}{x}}{\frac{1}{x}(x-1) + (\ln x)(1)} = \lim_{x \to 1} \frac{\frac{x-1}{x}}{\frac{x-1}{x} + \ln x}$ Multiplying the numerator and denominator by $x$ simplifies this to: $\lim_{x \to 1} \frac{x-1}{x-1 + x \ln x}$ This is still of the form $\frac{0}{0}$. Applying L'Hospital's Rule again: $\lim_{x \to 1} \frac{1}{1 + (1 \cdot \ln x + x \cdot \frac{1}{x})} = \lim_{x \to 1} \frac{1}{1 + \ln x + 1} = \lim_{x \to 1} \frac{1}{2 + \ln x} = \frac{1}{2+0} = \frac{1}{2}.$
6. What is the value of the limit $\lim_{x \to 1^+} (x-1) \tan\left(\frac{\pi x}{2}\right)$?
   Answer: $-\frac{2}{\pi}$ The limit is of the indeterminate form $0 \cdot (-\infty)$. To apply L'Hospital's Rule, we rewrite the expression as a quotient: $\lim_{x \to 1^+} \frac{x-1}{\cot\left(\frac{\pi x}{2}\right)}$ This is now of the form $\frac{0}{0}$. Applying L'Hospital's Rule: $\lim_{x \to 1^+} \frac{\frac{d}{dx}(x-1)}{\frac{d}{dx}\cot\left(\frac{\pi x}{2}\right)} = \lim_{x \to 1^+} \frac{1}{-\csc^2\left(\frac{\pi x}{2}\right) \cdot \frac{\pi}{2}}$ As $x \to 1^+$, $\frac{\pi x}{2} \to \frac{\pi}{2}$, and $\csc\left(\frac{\pi}{2}\right) = 1$. So the limit is: $\frac{1}{-1^2 \cdot \frac{\pi}{2}} = -\frac{2}{\pi}.$
7. Let $f$ be a twice-differentiable function such that $f(1)=0$, $f'(1)=0$, and $f''(1)=6$. Evaluate $\lim_{x \to 1} \frac{f(x)}{(x-1)^2}$.
   Answer: $3$ The limit is of the form $\frac{f(1)}{(1-1)^2} = \frac{0}{0}$. We apply L'Hospital's Rule: $\lim_{x \to 1} \frac{f'(x)}{2(x-1)}$ Since $f'(1)=0$, this is also of the form $\frac{0}{0}$. We apply L'Hospital's Rule a second time: $\lim_{x \to 1} \frac{f''(x)}{2}$ Since $f$ is twice-differentiable, $f''$ is continuous near 1, so we can substitute $x=1$: $\frac{f''(1)}{2} = \frac{6}{2} = 3.$
8. Evaluate $\lim_{x \to 0^+} (e^x + 2x)^{1/x}$.
   Answer: $e^3$ This limit is of the indeterminate form $1^\infty$. Let $L = \lim_{x \to 0^+} (e^x + 2x)^{1/x}$. We take the natural logarithm of both sides: $\ln(L) = \lim_{x \to 0^+} \frac{1}{x} \ln(e^x + 2x) = \lim_{x \to 0^+} \frac{\ln(e^x + 2x)}{x}$ This is of the form $\frac{0}{0}$. Applying L'Hospital's Rule: $\ln(L) = \lim_{x \to 0^+} \frac{\frac{e^x+2}{e^x+2x}}{1} = \frac{e^0+2}{e^0+0} = \frac{1+2}{1} = 3.$ Since $\ln(L) = 3$, the original limit is $L = e^3$.
9. Evaluate the limit $\lim_{x \to 0} \frac{1}{x^2} \int_0^x t \cos(t) \, dt$.
   Answer: $\frac{1}{2}$ First, rewrite the limit as a quotient: $\lim_{x \to 0} \frac{\int_0^x t \cos(t) \, dt}{x^2}$ At $x=0$, the numerator is $\int_0^0 t \cos(t) \, dt = 0$ and the denominator is $0^2=0$. The form is $\frac{0}{0}$. We apply L'Hospital's Rule. By the Fundamental Theorem of Calculus, Part 1, the derivative of the numerator is $x \cos(x)$. The derivative of the denominator is $2x$. $\lim_{x \to 0} \frac{x \cos(x)}{2x} = \lim_{x \to 0} \frac{\cos(x)}{2}$ Now, we can evaluate by direct substitution: $\frac{\cos(0)}{2} = \frac{1}{2}.$

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