Practice: Limits

Since directly substituting $x=0$ into the expression gives us $\frac {e^{44(0)}-1}{0^2+2(0)} = \frac {0}{0}$ , we can use L'Hopital's rule and differentiate the numerator and the denominator of the fraction.

$\lim_{x \to 0} {\frac {e^{44x}-1}{x^2+2x}} = \lim_{x \to 0} {\frac {44e^{44x}}{2x+2}} = \frac {44e^0}{2(0)+2} = \frac {44}{2} = \boxed{22}$

Is clear that the expression can be rewritten as $\frac{\text{e}^{44x}-1}{x^2+2x}=\frac{\text{e}^{44x}-1}{44x}\cdot\frac{44x}{x^2+2x}$ so by the fundamental limit $\lim_{t \to 0}\frac{\text{e}^t-1}{t} =1$ we have $\frac{\text{e}^{44x}-1}{44x}\cdot\frac{44x}{x^2+2x}=\frac{\text{e}^{44x}-1}{44x}\cdot\frac{44}{x+2} \to 1\cdot \frac{44}{2} =22$

This is a $\dfrac {0}{0}$ form, so it is of an indeterminate form. So we use L'Hôpital's rule: $\begin{aligned} \lim_{x \to 0} \left( \dfrac {e^{44x} - 1}{x^2 + 2x} \right) &= \lim_{x \to 0} \left( \dfrac {44e^{44x}}{2x+2} \right) \\&= \lim_{x \to 0} \left( \dfrac {22e^{44x}}{x+1} \right) \\&= \dfrac {22 \cdot e^0}{0+1} = \boxed {22}. \end{aligned}$

at first if you put $x=0$ in the expression you get $\frac{0}{0}$ form, then applying l'hopital's rule i.e. differentiatiating the numerator and denominator we get $\frac{44e^{44x}}{2x+2}$ , then put x=0 and you'll get $\frac {44}{2}$ i.e 22.

use L Hopitals rule just take derivative on numerator and denominator seperately and then apply limit remember donot use qoutient rule besta luck