# Use the method of your choice to evaluate the following limits. lim_{(x,y)rightarrow(0,1)}frac{ysin x}{x(y+1)}

Use the method of your choice to evaluate the following limits.
$\underset{\left(x,y\right)\to \left(0,1\right)}{lim}\frac{y\mathrm{sin}x}{x\left(y+1\right)}$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

aprovard
Given limits:
$\underset{\left(x,y\right)\to \left(0,1\right)}{lim}\frac{y\mathrm{sin}x}{x\left(y+1\right)}$
We have to evaluate the limit.
The limit is computed as follows:
$\underset{\left(x,y\right)\to \left(0,1\right)}{lim}\frac{y\mathrm{sin}x}{x\left(y+1\right)}=\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{x}\underset{y\to 1}{lim}\frac{y}{y+1}$
Substitutes the limits:
We know that: $\underset{\theta \to 0}{lim}\frac{\mathrm{sin}\theta }{\theta }=1$
$\underset{\left(x,y\right)\to \left(0,1\right)}{lim}\frac{y\mathrm{sin}x}{x\left(y+1\right)}=1×\frac{1}{1+1}$
$\underset{\left(x,y\right)\to \left(0,1\right)}{lim}\frac{y\mathrm{sin}x}{x\left(y+1\right)}=\frac{1}{2}$
Therefore, the value of $\underset{\left(x,y\right)\to \left(0,1\right)}{lim}\frac{y\mathrm{sin}x}{x\left(y+1\right)}$ is $\frac{1}{2}$
Jeffrey Jordon