$\lim _{x \rightarrow 0} \frac{x y^{x}}{y^{x}-1}$

View step-by-step help

Home

Math Problems

Calculus

Find limits of polynomials and rational functions

Full solution

\lim _{x \rightarrow 0} \frac{x y^{x}}{y^{x}-1}

Identify Indeterminate Form: Identify the indeterminate form of the limit. We need to determine the form of the limit as $x$ approaches $0$ . If we substitute $x = 0$ directly into the function, we get $(0 \cdot y^0)/(y^0 - 1)$ , which simplifies to $0/0$ , an indeterminate form.
Apply L'Hôpital's Rule: Apply L'Hôpital's Rule. $\newline$ Since we have an indeterminate form of $0/0$ , we can apply L'Hôpital's Rule, which states that if the limit of $f(x)/g(x)$ as $x$ approaches a value $c$ is $0/0$ or $\infty/\infty$ , then the limit is the same as the limit of $f'(x)/g'(x)$ as $x$ approaches $c$ , provided that the derivatives exist and $g'(x) \neq 0$ near $c$ .
Differentiate Numerator and Denominator: Differentiate the numerator and denominator with respect to $x$ . We need to find the derivatives of the numerator, $f(x) = xy^{x}$ , and the denominator, $g(x) = y^{x} - 1$ , with respect to $x$ . For the numerator, using the product rule and the chain rule, we get: $f'(x) = y^{x} + xy^{x}\ln(y)$ . For the denominator, using the chain rule, we get: $g'(x) = y^{x}\ln(y)$ . Now we can rewrite the limit using these derivatives.
Evaluate New Limit: Evaluate the new limit using the derivatives. $\newline$ The new limit is now: $\newline$ $\lim_{x \to 0}(y^{x} + xy^{x}\ln(y))/(y^{x}\ln(y))$ . $\newline$ We can now substitute $x = 0$ into this new expression: $\newline$ $\lim_{x \to 0}(y^0 + 0\cdot y^0\cdot \ln(y))/(y^0\cdot \ln(y)) = (1 + 0)/(\ln(y)) = 1/\ln(y)$ .
Check Final Expression: Check if the final expression is well-defined. $\newline$ Since $\ln(y)$ is undefined for $y \leq 0$ , we must assume $y > 0$ for the limit to exist. Given this assumption, the final expression $\frac{1}{\ln(y)}$ is well-defined, and we have successfully evaluated the limit.