IB Maths AA HL First Principles Questions

1. Test the limit by direct substitution of $x=0$: $\frac{e^0-1}{\sin 0}=\frac{1-1}{0}=\frac{0}{0}$.

2. State that because this evaluates to the indeterminate form $\frac{0}{0}$, L'Hôpital's Rule can be applied.

3. Differentiate the numerator with respect to $x$: $\frac{d}{dx}(e^{3x}-1)=3e^{3x}$.

4. Differentiate the denominator with respect to $x$: $\frac{d}{dx}(\sin x)=\cos x$.

5. Form the new limit fraction: $\underset{x\to 0}{lim}\frac{3e^{3x}}{\cos x}$.

6. Evaluate the new limit by substituting $x=0$: $\frac{3e^0}{\cos 0}=\frac{3(1)}{1}=\mathbf{3}$.

1. State the formal limit definition of the derivative: $f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$.

2. Substitute the specific function into the formula: $f^{\prime }(x)=\underset{h\to 0}{lim}\frac{(x+h{)}^3-x^3}{h}$.

3. Expand the binomial cubic correctly: $f^{\prime }(x)=\underset{h\to 0}{lim}\frac{x^3+3x^{2}h+3x{h}^2+h^3-x^3}{h}$.

4. Cancel the $x^3$ terms in the numerator: $f^{\prime }(x)=\underset{h\to 0}{lim}\frac{3x^{2}h+3x{h}^2+h^3}{h}$.

5. Factor out and cancel the common $h$ term: $f^{\prime }(x)=\underset{h\to 0}{lim}(3x^2+3xh+h^2)$.

6. Evaluate the limit by directly substituting $h=0$: $3x^2+3x(0)+(0{)}^2=\mathbf{3}{\mathbf{x}}^{\mathbf{2}}$.

1. Recognize that applying L'Hôpital's rule would require differentiating four times, so substituting the Maclaurin series for $\cos x$ is much faster.

2. State the standard Maclaurin expansion for $\cos x$: $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots$

3. Substitute this expansion into the numerator of the limit: $\underset{x\to 0}{lim}\frac{(1-\frac{x^2}{2}+\frac{x^4}{24}-\dots )-1+\frac{x^2}{2}}{x^4}$.

4. Cancel the constant $1$ and the $x^2$ terms perfectly: $\underset{x\to 0}{lim}\frac{\frac{x^4}{24}-\frac{x^6}{720}+\dots }{x^4}$.

5. Divide every term in the numerator by the denominator $x^4$: $\underset{x\to 0}{lim}(\frac{1}{24}-\frac{x^2}{720}+\dots )$.

6. Evaluate the limit by letting $x$ approach zero, annihilating all higher-order terms: $\frac{\mathbf{1}}{\mathbf{24}}$.